Binomial Theorem

1. BINOMIAL EXPRESSION

An algebraic expression consisting of only two terms separated by an operator {+, -} is called a Binomial Expression

Example:

a) x + 2y               b) 3x – 5y

c) (a + x)              d) $x^2+\frac{1}{a^2}$

2. BINOMIAL THEOREM FOR POSITIVE INTEGRAL INDEX

The formula by which any power of a binomial expression can be expanded in the form of a series is known as Binomial Theorem. This theorem was given by Sir Issac Newton.
The rule by which any power of binomial can be expanded is called the binomial theorem. If n is a positive integer and x, $y\in C$ then

$(\mathrm{x}+\mathrm{y}{)}^{\mathrm{n}}=\mathrm{C}_0^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}–0}{\mathrm{y}}^0+\mathrm{C}_1^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}–1}{\mathrm{y}}^1+\mathrm{C}_2^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}–2}{\mathrm{y}}^2$+……+$\mathrm{C}_{\mathrm{r}}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}–\mathrm{r}}{\mathrm{y}}^{\mathrm{r}}+\dots \dots +$$\mathrm{C}_{\mathrm{n}–1}^{\mathrm{n}}{\mathrm{xy}}^{\mathrm{n}–1}+\mathrm{C}_{\mathrm{n}}^{\mathrm{n}}{\mathrm{x}}^0{\mathrm{y}}^{\mathrm{n}}$

i.e., $(\mathrm{x}+\mathrm{y}{)}^{\mathrm{n}}=\sum _{\mathrm{r}=0}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}^{\mathrm{n}}.{\mathrm{x}}^{\mathrm{n}–\mathrm{r}}.{\mathrm{y}}^{\mathrm{r}}$      …..(i)

Here $\mathrm{C}_0^{\mathrm{n}},\mathrm{C}_1^{\mathrm{n}},\mathrm{C}_2^{\mathrm{n}},\dots ..\mathrm{C}_{\mathrm{n}}^{\mathrm{n}}$ are called binomial coefficients $\mathrm{C}_{\mathrm{r}}^{\mathrm{n}}=\frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}–\mathrm{r})!}\text{ for }0\le \mathrm{r}\le \mathrm{n}$.

Important Tips

The number of terms in the expansion of ${\left(x+y\right)}^n$ are (n + 1).

The expansion contains decreasing power of x and increasing power of y. The sum of the powers of x and y in each term is equal to n.

The binomial coefficients $C_0^n,C_1^n,C_2^n\dots \dots .$ equidistant from beginning and end are equal i.e., $C_r^n=C_{n–r}^n$.

${\left(x+y\right)}^n$ = Sum of odd terms + sum of even terms. 

3. SOME IMPORTANT EXPANSIONS

(1) Replacing y  by – y  in (i), we get,

$(x–y{)}^n=C_0^n{x}^{n–0}.y^0–C_1^n{x}^{n–1}.y^1+C_2^n{x}^{n–2}.y^2$……+$(–1{)}^r C_r^n{x}^{n–r}.y^r+\dots .+(–1{)}^{n}C_n^n{x}^0.y^{\eta }$

i.e.,  $(x–y{)}^n=\sum _{r=0}^n(–1{)}^r C_r^n{x}^{n–r}.y^r$    …..(ii)

The terms in the expansion of ${\left(x–y\right)}^n$ are alternatively positive and negative, the last term is positive or negative according
as n is even or odd.

(2) Replacing x by 1 and y by x in equation (i) we get,

$(1+x{)}^n=C_0^n{x}^0+C_1^n{x}^1+C_2^n{x}^2$+…..+$C_r^n{x}^r+\dots \dots +C_n^n{x}^n\text{ i.e., }(1+x{)}^n=\sum _{r=0}^{n}C_r^n{x}^r$

This is expansion of (1+x)ⁿ in ascending power of x.

(3) Replacing x by 1 and y by – x in (i) we get,

$(1–x{)}^n=C_0^n{x}^0–C_1^n{x}^1+C_2^n{x}^2–\dots \dots +(–1{)}^{r}C_r^n{x}^r$+…+ $(–1{)}^{n n}{C}_n{x}^n$i.e.,  

$(1–x{)}^n=\sum _{r=0}^n(–1{)}^{r}C_r^n{x}^r$

(4) $(x+y{)}^n+(x–y{)}^n$(x+y)ⁿ  – (x-y)^{n = $2\left[C_0^n{x}^n{y}^0+C_2^n{x}^{n⌞2}{y}^2+C_4^n{x}^{n–4}{y}^4+\dots \dots .\right]$}

and $(x+y{)}^n–(x–y{)}^n=2\left[C_1_n{x}^{n–1}{y}^1+C_3^n{x}^{n–3}{y}^3+C_5^n{x}^{n–5}{y}^5+\dots \dots .\right]$

(5) The coefficient of ${\left(\mathrm{r}+1\right)}^{\mathrm{th}}$ term in the expansion of $(1+x{)}^n\text{ is }C_r^n$.

(6)  The coefficient of $x^r$ in the expansion of $(1+x{)}^n\text{ is }C_r^n$.

Note  : If n  is odd, then $(x+y{)}^n+(x–y{)}^n$ and $(x+y{)}^n–(x–y{)}^n$, both have the same number of terms equal to $\left(\frac{n+1}{2}\right)$

If n is even, then $(x+y{)}^n+(x–y{)}^n$ has $\left(\frac{n}{2}+1\right)$ terms and $(x+y{)}^n–(x–y{)}^n$ has $\frac{n}{2}$ terms.  

Illustration -1

The expression ${\left(\sqrt{x}–\frac{1}{x}\right)}^4–{\left(\sqrt{x}+\frac{1}{x}\right)}^4$ can be simplified as

(A) $–8x^{\frac{3}{2}}\left(x^3–1\right)$           (B)  $–8x^{\frac{–5}{2}}\left(x^3+1\right)$         (C) 2(x² + 6x^–1 + x^–4)     (D) 2(x⁴+6x^–1 + x^–2)

Solution

B

The given expression is ${\left(\sqrt{x}–\frac{1}{x}\right)}^4–{\left(\sqrt{x}+\frac{1}{x}\right)}^4$.

This expression can be simplified by using Binomial Theorem. This can be done as

${\left(\sqrt{x}–\frac{1}{x}\right)}^4={\left\{\sqrt{x}+\left(–\frac{1}{x}\right)\right\}}^4$ = $C_0^4(\sqrt{x}{)}^4+C_1^4(\sqrt{x}{)}^3\left(\frac{–1}{x}\right)+C_2^4(\sqrt{x}{)}^2{\left(\frac{–1}{x}\right)}^2$+$\mathrm{C}_3^4\left(\sqrt{\mathrm{x}}\right){\left(\frac{–1}{\mathrm{x}}\right)}^3+\mathrm{C}_4^4{\left(\frac{–1}{\mathrm{x}}\right)}^4$

= $x^2–4x^{\frac{1}{2}}+6x^{–1}–4x^{\frac{–5}{2}}+x^{–4}$

Similarly, it can be shown that,  

${\left(\sqrt{x}+\frac{1}{x}\right)}^4=x^2+4x^{\frac{1}{2}}+6x^{–1}+4x^{\frac{–5}{2}}+x^{–4}$

$\therefore {\left(\sqrt{x}–\frac{1}{x}\right)}^4–{\left(\sqrt{x}+\frac{1}{x}\right)}^4$

$\left(x^2+4x^{\frac{1}{2}}+6x^{–1}–4x^{\frac{–5}{2}}+x^{–4}\right)–\left(x^2+4x^{\frac{1}{2}}+6x^{–1}+4x^{\frac{–5}{2}}+x^{–4}\right)$

= $–8x^{\frac{1}{2}}–8x^{\frac{–5}{2}}=–8x^{\frac{–5}{2}}\left(x^3+1\right)$

Illustration -2

The expression[x¹⁰ – 5x⁷ + 10x⁴ – 10x + 5x^–2 – x^–5] is the expanded form the expression

(A) ${\left(x^5–\frac{1}{5}\right)}^2$       (B) ${\left(\frac{1}{5}–x^2\right)}^5$       (C) ${\left(x^2–\frac{1}{5}\right)}^5$      (D) ${\left(\frac{1}{5}+x^5\right)}^2$

Solution.

C

The given expression is x¹⁰ – 5x⁷ + 10x⁴ – 10x + 5x^–2 – x^–5.

x¹⁰ – 5x⁷ + 10x⁴ – 10x + 5x^–2 – x^–5

= $\mathrm{C}_0^5{\left({\mathrm{x}}^2\right)}^5+\mathrm{C}_1^5{\left({\mathrm{x}}^2\right)}^4\left(\frac{–1}{\mathrm{x}}\right)+\mathrm{C}_2^5{\left({\mathrm{x}}^2\right)}^3{\left(\frac{–1}{\mathrm{x}}\right)}^2$ + $\mathrm{C}_3^5{\left({\mathrm{x}}^2\right)}^2{\left(\frac{–1}{\mathrm{x}}\right)}^3+\mathrm{C}_4^5\left({\mathrm{x}}^2\right){\left(\frac{–1}{\mathrm{x}}\right)}^4+\mathrm{C}_5^5{\left(\frac{–1}{\mathrm{x}}\right)}^5$

= ${\left\{x^2+\left(\frac{–1}{5}\right)\right\}}^5={\left(x^2–\frac{1}{5}\right)}^5$

Thus, the given expression is the expanded form of the expression ${\left(x^2–\frac{1}{5}\right)}^5$.

Illustration -3

What is the expanded form of the expression ${\left(3x+\frac{1}{4y}\right)}^4$?

(A) $81x^4+\frac{27x^3}{y}+\frac{27x^2}{8y^2}+\frac{3x}{16y^3}+\frac{1}{256y^4}$          (B) $81x^4+\frac{27x^3}{8y}+\frac{9x^2}{16y^2}+\frac{3x}{4y^3}+\frac{1}{256y^4}$

(C) $324x^4+\frac{27x^3}{y}+\frac{27x^2}{8y^2}+\frac{3x}{16y^3}+\frac{1}{64y^4}$         (D) $324x^4+\frac{81x^3}{4y}+\frac{27x^2}{8y^2}+\frac{3x}{64y^3}+\frac{1}{64y^4}$

Solution

A

By using Binomial Theorem, the expression ${\left(3x+\frac{1}{4y}\right)}^4$ can be expanded as

${\left(3x+\frac{1}{4y}\right)}^4$ = ⁴C₀(3x)⁴ + ⁴C₁(3x)³ $\left(\frac{1}{4y}\right)$+ ⁴C₂(3x)² ${\left(\frac{1}{4y}\right)}^2$+ ⁴C₃(3x)${\left(\frac{1}{4y}\right)}^3$+ ⁴C₄ ${\left(\frac{1}{4y}\right)}^4$

= 3⁴x⁴ + 4.(3³x³)$\left(\frac{1}{4y}\right)$+ 6.(3²x²)$\left(\frac{1}{4^3{y}^2}\right)$+ 4.(3x)$\left(\frac{1}{4^3{y}^3}\right)+\frac{1}{4^4{y}^4}$  

= $81x^4+\frac{27x^3}{y}+\frac{27x^2}{8y^2}+\frac{3x}{16y^3}+\frac{1}{256y^4}$

Illustration -4

Expand  ${\left(x^2+\frac{3}{x}\right)}^4,x\ne 0$.

Solution

By using binomial theorem, we have

${\left(x^2+\frac{3}{x}\right)}^4=C_0^4{\left(x^2\right)}^4+C_1^4{\left(x^2\right)}^3\left(\frac{3}{x}\right)+C_2^4{\left(x^2\right)}^2{\left(\frac{3}{x}\right)}^2+C_3^4\left(x^2\right){\left(\frac{3}{x}\right)}^3+C_4^4{\left(\frac{3}{x}\right)}^4$

= $x^8+4x^6.\frac{3}{x}+6.x^4.\frac{9}{x^2}+4.x^2.\frac{27}{x^3}+\frac{81}{x^4}$ = $x^8+12x^5+54x^2+\frac{108}{x}+\frac{81}{x^4}$

Illustration -5

Using  binomial theorem, expand  $\left\{(x+y{)}^5+(x–y{)}^5\right\}$ and hence  find the value of  $\left\{(\sqrt{2}+1{)}^5+(\sqrt{2}–1{)}^5\right\}$.

Solution

We have, $(x+y{)}^5+(x–y{)}^5$ = $2\left[C_0^5{x}^5+C_2^5{x}^3{y}^2+C_4^5{x}^1{y}^4\right]=2\left(x^5+10x^3{y}^2+5x{y}^4\right)$

Putting $x=\sqrt{2}$  and y = 1, we get  

$(\sqrt{2}+1{)}^5+(\sqrt{2}–1{)}^5=2\left[(\sqrt{2}{)}^5+10(\sqrt{2}{)}^3+5\sqrt{2}\right]$ = $2[4\sqrt{2}+20\sqrt{2}+5\sqrt{2}]=58\sqrt{2}$

Illustration -6

Compute (98)⁵.

Solution

We express 98 as the sum or difference of two numbers whose powers are easier to calculate, and then use Binomial Theorem.

Write 98 = 100 – 2

Therefore, (98)⁵ = (100 – 2)⁵= ⁵C₀ (100)⁵ – ⁵C₁ (100)⁴.2 + ⁵C₂ (100)³2² – ⁵C₃ (100)² (2)³ + ⁵C₄ (100) (2)⁴ – ⁵C₅ (2)⁵

= 10000000000 – 5 × 100000000 × 2 + 10 × 1000000 × 4 – 10 ×10000 × 8 + 5 × 100 × 16 – 32

= 10040008000 – 1000800032 = 9039207968.

Illustration -7

Which is larger (1.01)^1000000 or 10,000?

Solution

Splitting 1.01 and using binomial theorem to write the first few terms we have (1.01)^1000000 = (1 + 0.01)^1000000

= ^1000000C₀ + ^1000000C₁(0.01) + other positive terms

= 1 + 1000000 × 0.01 + other positive terms = 1 + 10000 + other positive terms > 10000

Hence (1.01)^1000000 > 10000

Illustration -8

If, $(\sqrt{5}+\sqrt{3}{)}^4+(\sqrt{5}–\sqrt{3}{)}^5=a+b\sqrt{15}+c\sqrt{5}+d\sqrt{3}$ where a, b, c and d are integers, then what is the value of the expression a + b + c − d?

(A) – 348                 (B) – 156                  (C) 220                  (D) 660

Solution

D

By using binomial theorem, we obtain

$(\sqrt{5}+\sqrt{3}{)}^4=\sum _{r=0}^4\mathrm{C}_r^4(\sqrt{5}{)}^{4–r}(\sqrt{3}{)}^r,(\sqrt{5}–\sqrt{3}{)}^5=\sum _{\mathrm{p}}^5\mathrm{C}_{\mathrm{p}}^5(\sqrt{5}{)}^{5–\mathrm{p}}·(–\sqrt{3}{)}^{\mathrm{p}}$

∴ $(\sqrt{5}+\sqrt{3}{)}^4=\mathrm{C}_0^4(\sqrt{5}{)}^4+\mathrm{C}_1^4\left(\sqrt{5}{)}^3\right(\sqrt{3})+\mathrm{C}_2^4(\sqrt{5}{)}^2(\sqrt{3}{)}^2+\mathrm{C}_3^4\sqrt{5}(\sqrt{3}{)}^2+\mathrm{C}_4^4(\sqrt{3}{)}^4$

= $25+4.5\sqrt{5}·\sqrt{3}+6.5.3+4.\sqrt{5}.3\sqrt{3}+9=25+20\sqrt{15}+90+12\sqrt{15}+9=124+32\sqrt{15}$

$(\sqrt{5}–\sqrt{3}{)}^5=C_0^5(\sqrt{5}{)}^5+C_1^5\left(\sqrt{5}{)}^4\right(–\sqrt{3})+C_2^5(\sqrt{5}{)}^3(–\sqrt{3}{)}^2+C_3^5(\sqrt{5}{)}^2(–\sqrt{3}{)}^3$$+C_4^5\left(\sqrt{5}\right)(–\sqrt{3}{)}^4+C_5^5(–\sqrt{3}{)}^5$

= $1\times 25\sqrt{5}+5\left(25\right)(–\sqrt{3})+10\left(5\sqrt{5}\right)\left(3\right)+10.5·(–3\sqrt{3})+5·\left(\sqrt{5}\right)\left(9\right)+1·(–9\sqrt{3})$

= $25\sqrt{5}–125\sqrt{3}+150\sqrt{5}–150\sqrt{3}+45\sqrt{5}–9\sqrt{3}=220\sqrt{5}–284\sqrt{3}$

∴ $(\sqrt{5}+\sqrt{3}{)}^4+(\sqrt{5}–\sqrt{3}{)}^5=124+32\sqrt{15}+220\sqrt{5}–284\sqrt{3}$, which is of the form $a+b\sqrt{15}+c\sqrt{5}+d\sqrt{3}$,

where a = 124, b = 32, c = 220 and d = −284

Thus, a + b + c − d = 124 + 32 + 220 − (− 284) = 124 + 32 + 220 + 284 = 660

4. General term in a Binomial expansion

GENERAL TERM

(a + x)ⁿ = ⁿC₀aⁿx₀ + ⁿC₁a^{n – 1}x¹ + ⁿC₂a^{n – 2}x² +…. + ⁿC_na⁰ xⁿ

Here we see that the first term    = ⁿC₀aⁿ x⁰ ,

Second term                                     = ⁿC₁a^{n – 1}x¹,

Third term                                        = ⁿC₂a^{n – 2}x²,

Fourth term                                      = ⁿC₃a^{n – 3}x³,

                                                            ………………….

Similarly, r^th term                           = ⁿC_{r – 1} a^{n – (r –1)} x^{r – 1} ,

If in the expansion of (a + x)ⁿ, r^th term is denoted by T_r, then T_r = ⁿC_{r – 1} a^{n –r +1} x^{r – 1}

r^th term is called the general term.

(r + 1)^th term may also be called  the general term.

(r + 1)^th term in the expansion of (a + x)ⁿ is given by    T_r+1 = ⁿC_r a^{n –r}x^r  

• In the binomial expansion of $(x–y{)}^n,T_{r+1}=(–1{)}^{r}C_r^n{x}^{n–r}{y}^r$
• In the binomial expansion of $(1+x{)}^n,T_{r+1}=C_r^n{x}^r$
• In the binomial expansion of $(1–x{)}^n,T_{r+1}=(–1{)}^r C_r^n{x}^r$
• In the binomial expansion of (x + y)ⁿ, the p^th term from the end is term from beginning.

Important Tips

In the expansion of $(x+y{)}^n, n\in N$

$\frac{T_{r+1}}{T_r}=\left(\frac{n–r+1}{r}\right)\frac{y}{x}$

The coefficient of $x^{n–1}$ in the expansion of $(x–1)(x–2)\dots \dots .(x–n)=–\frac{n(n+1)}{2}$

The coefficient of  in the expansion of  $(x+1)(x+2)\dots ..(x+n)=\frac{n(n+1)}{2}$

Illustration -9

If the 4th term in the expansion of  ${\left(\mathrm{px}+\frac{1}{\mathrm{x}}\right)}^{\mathrm{m}}\text{ is }\frac{5}{2}\forall \mathrm{x}\in \mathrm{ℝ}$.

Solution

Since,  ${\mathrm{t}}_4={\mathrm{t}}_{3+1}=\frac{5}{2}$  (given)

$\Rightarrow C_3^m(px{)}^{m–3}{\left(\frac{1}{x}\right)}^3=\frac{5}{2}$

$\Rightarrow \frac{m(m–1)(m–2)}{3!}{p}^{m–3}{x}^{m–6}=\frac{5}{2}$

Since, the term can be seen is independent of x, hence,m = 6

Now, for m = 6, we have

$\frac{6(6–1)(6–2)}{6}{p}^3=\frac{5}{2}$

$\therefore \boxed{p=\frac{1}{2}}$

Illustration -10

Find the coefficient of $x^9$ in the expansion of ${\left(x^2–\frac{1}{3x}\right)}^9$.

Solution

General term in the expansion of ${\left(x^2–\frac{1}{3x}\right)}^9$ is $t_{r+1}=C_r^9{\left(x^2\right)}^{9–r}{\left(–\frac{1}{3x}\right)}^r$

or $t_{r+1}=\frac{(–1{)}^r}{3^r}C_r^9(x{)}^{18–3r}$

For coefficient of  $x^9,18–3r=9$  i.e. r = 3

${\mathrm{t}}_4=–\frac{1}{27}\mathrm{C}_3^9=–\frac{28}{9}$

Illustration -11

Find the 11th term in expansion of ${\left(3x–\frac{1}{x\sqrt{3}}\right)}^{20}$.

Solution

The general term $t_{r+1}=(–1{)}^{r^{20}}{C}_r(3x{)}^{20–r}{\left(\frac{1}{x\sqrt{3}}\right)}^r$

For the  11th term , we must  take r = 10

$\therefore t_{11}=t_{10+1}=(–1{)}^{10}C_{10}^{20}(3x{)}^{20–10}{\left(\frac{1}{x\sqrt{3}}\right)}^{10}$

= $\mathrm{C}_{10}^{20} 3^{10}{\mathrm{x}}^{10}\frac{1}{{\mathrm{x}}^{10}(\sqrt{3}{)}^{10}}=\mathrm{C}_{10}^{20}(\sqrt{3}{)}^{10}=\mathrm{C}_{10}^{20}{3}^5$

Illustration -12

Find  the coefficient of $x^{32}$ and $x^{–17}$ in the  expansion of  ${\left(x^4–\frac{1}{x^3}\right)}^{15}$.

Solution

Suppose  (r + 1)th term involves  $x^{32}$ in the expansion of ${\left(x^4–\frac{1}{x^3}\right)}^{15}$

Now,  ${\mathrm{T}}_{\mathrm{r}+1}=\mathrm{C}_{\mathrm{r}}^{15}{\left({\mathrm{x}}^4\right)}^{15–\mathrm{r}}{\left(–\frac{1}{{\mathrm{x}}^3}\right)}^{\mathrm{r}}=(–1{)}^{\mathrm{r}15}{\mathrm{C}}_{\mathrm{r}}{\mathrm{x}}^{60–7\mathrm{r}}$  ….(i)

For this  term to contain $x^{32}$, we must  have $60–7r=32\Rightarrow r=4$.

So, (4 + 1)th i.e. 5th term contains $x^{32}$.

Putting r = 4 in (i) , we get $T_5=(–1{)}^{4}C_4^{15}{x}^{(60–28)}=C_4^{15}{x}^{32}$

Coefficient of ${\mathrm{x}}^{32}=\mathrm{C}_4^{15}=1365$

Suppose (s + 1)th term in the binomial expansion of ${\left(x^4–\frac{1}{x^3}\right)}^{15}$ contains $x^{–17}$

Now,  ${\mathrm{T}}_{\mathrm{s}+1}=\mathrm{C}_{\mathrm{s}}^{15}{\left({\mathrm{x}}^4\right)}^{15–\mathrm{s}}{\left(–\frac{1}{{\mathrm{x}}^3}\right)}^{\mathrm{s}}=(–1{)}^{\mathrm{s}15}{\mathrm{C}}_{\mathrm{s}}{\mathrm{x}}^{60–7 \mathrm{s}}$….(ii)

If this term contains $x^{–17}$, we  must have $60–7s=–17\Rightarrow s=11$

So, (11+1)th i.e. 12th term contains $x^{–17}$.

Putting  s = 11 in (ii), we get

$T_{12}=(–1{)}^{{11}^{15}}{C}_{11}{x}^{–17}=–C_{11}^{15}{x}^{–17}=–C_4^{15}{x}^{–17} \left[\because C_r^n=C_{n–r}^n\right]$

Coefficient of $x^{–17}=–\mathrm{C}_4^{15}=–1365$.  

Illustration -13

Find the coefficient $x^5$ of  in the expansion of the product $(1+2x{)}^6(1–x{)}^7$.

Solution

We have   $(1+2x{)}^6(1–x{)}^7=\left[1+\mathrm{C}_1^6\left(2x\right)+C_2^6(2x{)}^2+C_3^6(2x{)}^3+C_4^6(2x{)}^4+C_5^6(2x{)}^5\right.$

= $\left(1+12x+60x^2+160x^3+240x^4+192x^5+\dots ..\right)\times$$\left(1–7x+21x^2–35x^3+35x^4–21x^5+\dots \right)$

∴ Coefficient of $x^5$ in the product

= $1\times (–21)+12\times 35+60\times (–35)+160\times 21+240\times (–7)+192\times 1$

= $–21+420–2100+3360–1680+192=171$

Illustration -14

If the coefficients of the eighth and ninth terms in the binomial expansion of are equal, then what is the value of n?

(A) 25            (B) 40            (C) 50            (D) 55

Solution

D

The general term, T_{r + 1} in the binomial expansion of ${\left(2+\frac{x}{3}\right)}^n$ is given by

T_r+1 = ⁿC_r(2)^{n – r} ${\left(\frac{\mathrm{x}}{3}\right)}^r$.

For r = 7, T₈ = ⁿC₇(2)^{n – 7} ${\left(\frac{\mathrm{x}}{3}\right)}^r$

∴ Coefficient of T₈ = ⁿC₇(2)^{n – 7} .$\frac{1}{3^7}$

For r = 8, T₉ = ⁿC₈(2)^{n – 8} ${\left(\frac{\mathrm{x}}{3}\right)}^8$

∴ Coefficient of T₉ = ⁿC₈(2)^{n – 8.}  $\frac{1}{3^8}$

Coefficient of T₈ = Coefficient of T₉

ⁿC₇.(2)^{n – 7.} $\frac{1}{3^7}$= ⁿC₈.(2)^{n – 8.} $\frac{1}{3^8}$

$\Rightarrow \frac{\mathrm{n}!}{7!(\mathrm{n}–7)!}\times 2=\frac{\mathrm{n}!}{8!(\mathrm{n}–8)!}·\frac{1}{3}\Rightarrow \frac{6}{\mathrm{n}–7}=\frac{1}{8}=$

⇒ n – 7 = 48Þ n = 48 + 7 = 55

Thus, the required value of n is 55.

5. INDEPENDENT TERM OR CONSTANT TERM

Independent term or constant term of a binomial expansion is the term in which exponent of the variable is zero.

Condition:  [Power of x] + r . [Power of y] = 0, in the expansion of ${\left[x+y\right]}^n$.

Illustration -15

What is the ratio of the coefficient of x¹⁰ in the expansion of (1 − x²)¹⁰ to the term independent of x in the expansion of ${\left(x–\frac{2}{x}\right)}^{10}$?

(A) 1:8             (B) 1:16            (C) 1:32            (D) 1:64

Solution

C

The (r + 1)^th term, T_{r + 1}, in the expansion of (1 − x²)¹⁰ is given by,

T_r+1 = ¹⁰C_r(– 1)^r (x2)^r

Putting r = 5, we obtain

T₆ = –¹⁰C₅x¹⁰

∴ Coefficient of x¹⁰ = –¹⁰C₅

The (r + 1)^th term, T_{r + 1}, in the expansion of ${\left(x–\frac{2}{x}\right)}^{10}$ is given by,

T_r+1 = ¹⁰C_r(– 1)^r x^10–r ${\left(\frac{2}{x}\right)}^2$= (–1)^{r 10}C_r2^rx^10–2r

Putting 10 − 2r = 0, we obtain

Term independent of x, T₆ = (–1)^{5 10}C₅.2⁵ = – 32¹⁰C₅

Hence, required ratio = (–¹⁰C₅):(–32¹⁰C₅) = 1:32

Illustration -16

Which term is independent of x in the binomial expansion of ${\left(\frac{4x}{5}–\frac{1}{4x^3}\right)}^4$?

(A) $–\frac{64}{125}$        (B) $–\frac{16}{125}$         (C) $\frac{4}{125}$           (D) $\frac{8}{125}$

Solution

A

The given expansion is ${\left(\frac{4x}{5}–\frac{1}{4x^3}\right)}^4$.

∴ General term, T_r+1 = (–1)^r = ⁴C_r  ${\left(\frac{4x}{5}\right)}^{4–r}{\left(\frac{1}{4x^3}\right)}^r=(–1{)}^{r^4}{C}_r{\left(\frac{4}{5}\right)}^{4–r}·\frac{1}{4^r}{x}^{4–4r}$

The term is independent of x if 4 − 4r = 0.

4 − 4r = 0

⇒ r = 1

Thus, term independent of x = (– 1)^{1 4}C₁ ×  ${\left(\frac{4}{5}\right)}^{(4–1)}\times \left(\frac{1}{4}\right)=–4\times \frac{64}{125}\times \frac{1}{4}=–\frac{64}{125}$

Illustration -17

Which term is independent of p in the expansion of ${\left(\frac{5}{2p}–8p^3\right)}^8$?

(A) second          (B) third            (C) fourth       (D) fifth  

Solution

B

It is known that the general term, T_r+1 i.e. (r + 1)^th term, of the binomial expansion of (a + b)ⁿ is given by,

T_r+1= ⁿC_ra^n–rb^r

Therefore, for the expansion of, ${\left(\frac{5}{2p}–8p^3\right)}^8$we have

$T_{r+1}=C_r^8{\left(\frac{5}{2p}\right)}^{8–r}{\left(–8p^3\right)}^r=C_r^8{\left(\frac{5}{2}\right)}^{8–r}{\left(\frac{1}{p}\right)}^{8–r}$ (–1)^r(8)^r(p)^3r 

= $(–1{)}^{r8}{C}_r\frac{(5{)}^{8–r}{\left(2^3\right)}^r}{(2{)}^{8–t}}·\frac{(p{)}^{3r}}{(p{)}^{8–r}}$

= $(–1{)}^{r{ }^8}{\mathrm{C}}_{\mathrm{r}}\frac{(5{)}^{8–r}}{(2{)}^{8–4r}}·(\mathrm{p}{)}^{4r–8}$

The term will be independent of p if the index of p is 0.

4r − 8 = 0

⇒ 4r = 8 ⇒ r = 2

Thus, the third term of the expansion of ${\left(\frac{5}{2p}–8p^3\right)}^8$ is independent of p.

Illustration -18

Which term is independent of the variable a in the expansion of ${\left(4a–\frac{1}{a^{\frac{1}{3}}}\right)}^{16}, a\ne 0$?

(A) 11^th             (B) 12^th                         (C) 13^th             (D) 14^th

Solution

C

It is known that the general term T_r+1 {which is the (r + 1)^th term} in the binomial expansion of

(a + b)ⁿ is given by T_r+1= ⁿC_r a^n–rb^r

Therefore, the general term in the expansion of ${\left(4a–\frac{1}{a^{\frac{1}{3}}}\right)}^{16}$ is

T_r+1 = ¹⁶C_r(4a)^{16 – r} ${\left(–a^{\frac{1}{3}}\right)}^r$

= (–1)^r.¹⁶C_r(4)^16–r.(a)^16–r. $a^{\frac{r}{3}}$  = (–1)^r.¹⁶C_r(4)^16–r. $a^{16–r–\frac{r}{3}}$     __________ (1)

This term will be independent of a, if $16–r–\frac{r}{3}=0$

$\Rightarrow 16=r+\frac{r}{3}\Rightarrow 16=\frac{4}{3}r$     $\Rightarrow r=\frac{16\times 3}{4}\Rightarrow r=12$

Thus, the 13^th term is independent of a in the expansion of ${\left(4a–\frac{1}{a^{\frac{1}{3}}}\right)}^{16}$.

Illustration -19

Find the term independent of x in the expansion of ${\left(3x^2–\frac{1}{2x^3}\right)}^{10}$.

Solution

Let  (r + 1)^th  term  be independent   of x in the given expression.

Now,  ${\mathrm{T}}_{\mathrm{r}+1}=\mathrm{C}_{\mathrm{r}}^{10}{\left(3{\mathrm{x}}^2\right)}^{10–\mathrm{r}}{\left(–\frac{1}{2{\mathrm{x}}^3}\right)}^{\mathrm{r}}=\mathrm{C}_{\mathrm{r}}^{10}{3}^{10–r}{\left(–\frac{1}{2}\right)}^{\mathrm{r}}{\mathrm{x}}^{20–5\mathrm{r}}$  …(i)

This term is independent   of x, if  $20–5r=0\Rightarrow r=4$.

So, (4 +1)th i.e. 5th term is independent   of x. Putting   r = 4 in (i), we get  

${\mathrm{T}}_5=\mathrm{C}_4^{10}{3}^6{\left(–\frac{1}{2}\right)}^4=\frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}\times \frac{729}{16}=\frac{76545}{8}$

Hence, required term = $\frac{76545}{8}$.

6. Middle term in a Binomial expansion

The middle term depends upon the value of n.

(1) When n is even, then total number of terms in the expansion of (x+y)ⁿ is n+1 (odd). So there is only one middle term i.e., ${\left(\frac{n}{2}+1\right)}^{th}$ term is the middle term. $T_{\left[\frac{n}{2}+1\right]}=C_{n/2}^n{x}^{n/2}{y}^{n/2}$

(2) When n is odd, then total number of terms in the expansion of (x+y)ⁿ is n+1 (even). So, there are two middle terms i.e., ${\left(\frac{n+1}{2}\right)}^{\text{th }}\text{ and }{\left(\frac{n+3}{2}\right)}^{\text{th }}$ are two middle terms.

$T_{\left(\frac{n+1}{2}\right)}=C_{\frac{n–1}{2}}^n{x}^{\frac{n+1}{2}}{y}^{\frac{n–1}{2}}$  and   $T_{\left(\frac{n+3}{2}\right)}=C_{\frac{n+1}{2}}^n{x}^{\frac{n–1}{2}}{y}^{\frac{n+1}{2}}$

Note :  When there are two middle terms in the expansion then their binomial coefficients are equal. Binomial coefficient of middle term is the greatest binomial coefficient.  

Illustration -20

Find the middle terms in the expansions of ${\left(3–\frac{x^3}{6}\right)}^7$

Solution

It is known that in the expansion of (a + b)ⁿ, if n is odd, then there are two middle terms, namely, ${\left(\frac{n+1}{2}\right)}^{\text{th }}$ term and ${\left(\frac{n+1}{2}+1\right)}^{th}$ term.

Therefore, the middle terms in the expansion of  ${\left(3–\frac{x^3}{6}\right)}^7\text{ are }{\left(\frac{7+1}{2}\right)}^{\text{th }}=4^{\text{th }}$ term and       ${\left(\frac{7+1}{2}+1\right)}^{\text{th }}=5^{\text{th }}$ term

T₄ = T₃₊₁ = ⁷C₃(3)^{7 – 3} ${\left(–\frac{x^3}{6}\right)}^3$= (–1)³  $\frac{7!}{3!4!}·3^4·\frac{x^9}{6^3}$

= $–\frac{7.6.5.4!}{3.2.4!}·3^4·\frac{1}{2^3·3^3}·x^9=–\frac{105}{8}{x}^9$

T₅ = T₄₊₁ = ⁷C₄(3)^{7 – 4} ${\left(–\frac{x^3}{6}\right)}^3=(–1{)}^4\frac{7!}{4!3!}·(3{)}^3·\frac{x^{12}}{6^4}$  

$=\frac{7.6.5.4!}{4!.3.2}·\frac{3^3}{2^4·3^4}·x^{12}=\frac{35}{48}{x}^{12}$

Thus, the middle terms in the expansion of ${\left(3–\frac{x^3}{6}\right)}^7\text{ are }–\frac{105}{8}{x}^9\text{ and }\frac{35}{48}{x}^{12}$.

Illustration -21

Find the middle terms in the expansions of ${\left(\frac{x}{3}+9y\right)}^{10}$

Solution

It is known that in the expansion (a + b)ⁿ, if n is even, then the middle term is ${\left(\frac{\mathrm{n}}{2}+1\right)}^{\text{th }}$ term.

Therefore, the middle term in the expansion of ${\left(\frac{x}{3}+9y\right)}^{10}$ is ${\left(\frac{10}{2}+1\right)}^{th}=6^{th}$ term

T₆ = T₅₊₁ = ¹⁰C₅  ${\left(\frac{x}{3}\right)}^{10–5}(9y{)}^5=\frac{10!}{5!5!}·\frac{x^5}{3^5}·9^5·y^5$

= $\frac{10.9·8·7·6·5!}{5·4·3·2·5!}·\frac{1}{3^5}·3^{10}·x^5{y}^5$      [9⁵ = (3²)⁵ = 3¹⁰]

252 × 3⁵ . x⁵ . y⁵ = 61236x⁵y⁵

Thus, the middle term in the expansion of ${\left(\frac{x}{3}+9y\right)}^{10}$ is 61236 x⁵y⁵.

Illustration -22

What is the middle term in the expansion of (1 − 2x + x²)ⁿ?

(A) $\frac{\left(2n\right)!}{n!n!}·x^n$     (B) $\frac{\left(2n\right)!}{n!n!}(–x{)}^n$    (C) $\frac{\left(2n\right)!}{n!}(–x{)}^n$    (D) $\frac{\left(2n\right)!}{n!}·x^n$

Solution

B

We have (1 – 2x + x²)ⁿ = [(1 – x)²]ⁿ = (1 – x)²ⁿ

Here, 2ⁿ is an even integer.

Therefore, ${\left(\frac{2n}{2}+1\right)}^{\text{th }}$ term i.e., (n + 1)^th term is the middle term.

(n + 1)^th term in the expansion of (1 – x)²ⁿ

= ²ⁿC_n(1)^2n–n(–x)ⁿ = ²ⁿC_n(–x)ⁿ = $\frac{\left(2n\right)!}{n! n!}(–x{)}^n$

Illustration -23

What is the middle term in the binomial expansion of (x⁴ − 4x⁵ + 6x⁶ − 4x⁷ + x⁸)⁵?

(A) $–\mathrm{C}_{10}^{20}{\mathrm{x}}^{30}$    (B) $–\mathrm{C}_{10}^{20}\frac{1}{{\mathrm{x}}^{30}}$    (C)  $\mathrm{C}_{10}^{20}\frac{1}{{\mathrm{x}}^{30}}$    (D) $\mathrm{C}_{10}^{20}{\mathrm{x}}^{30}$

Solution

D

x⁴ – 4x⁵ + 6x⁶ – 4x⁷ + x⁸ = x⁴(1 – 4x + 6x² – 4x³ + x⁴)

= x⁴(⁴C₀ – ⁴C₁x + ⁴C₂x² – ⁴C₃x³ + ⁴C₄x⁴)

= x⁴(1 – x)⁴ $\left[(1–a{)}^n=\sum _{r=0}^{1}C_r^n(–a{)}^r\right]$

= (x – x²)⁴

∴ (x⁴ – 4x⁵ + 6x⁶ – 4x⁷ + x⁸)⁵ = [(x – x²)⁴]⁵ = (x – x²)²⁰

Hence, the middle terms in the expansions of (x⁴ − 4x⁵ + 6x⁶ − 4x⁷ + x⁸)⁵ and (x − x²)²⁰ are equal.

Consider the expansion of (x − x²)²⁰.

Here, n = 20, which is even.

∴ Middle term term = ${\left(\frac{20}{2}+1\right)}^{\mathrm{th}}={11}^{\text{th }}\text{ term }$

Now, general term, t_r+1 = (–1)^{r 20}C_r(x)^{20 – r}(x²)^r = (– 1)^{r 20}C_rx^20+r

For r = 10, t₁₁ = (–1)^{10 20}C₁₀x³⁰ = ²⁰C₁₀x³⁰

Thus, the middle term in the binomial expansion of the given expression is ²⁰C₁₀x³⁰.

Illustration -24

What is the middle term in the expansion of (3x − y)¹⁰?

(A) 306180x⁵y⁵     (B) 61236x⁵y⁵   (C) –61236x⁵y⁵      (D) –306180x⁵y⁵

Solution

C

It is known that in the expansion of (a + b)ⁿ, if n is even, then the middle term is ${\left(\frac{n}{2}+1\right)}^{\text{th }}$ term.

Therefore, the middle term in the expansion of (3x − y)¹⁰ is ${\left(\frac{10}{2}+1\right)}^{\text{th }}=6^{\text{th }}$ term

6^th term in the expansion of (3x − y)¹⁰ is given by,

T₆ = T₅₊₁ = ¹⁰C₅(3x)^10–5(–y)⁵ =.(3⁵x⁵).y⁵  = $–\frac{10!}{5! 5!}·\left(3^5{\mathrm{x}}^5\right)·{\mathrm{y}}^5$

= $–\frac{10.9.8.7.6.5!}{5.4.3.2.5!}·3^5·x^5{y}^5$

= –252 × 3⁵.x⁵y⁵ = –61236x⁵y⁵

Thus, the middle term in the expansion of (3x − y)¹⁰ is –61236x⁵y⁵.

7. Determination of a particular term in the expansion

In the expansion of ${\left(x^{\alpha }\pm \frac{1}{x^{\beta }}\right)}^n$, if $x^m$ occurs in $T_{r+1}$, then r is given by $n\alpha –r(\alpha +\beta )=m\Rightarrow r=\frac{n\alpha –m}{\alpha +\beta }$

Thus in above expansion if constant term which is independent of x, occurs in $T_{r+1}$ then r is determined by

$n\alpha –r(\alpha +\beta )=0\Rightarrow r=\frac{n\alpha }{\alpha +\beta }$

Illustration -25

The 3rd, 4th and 5th terms in the expansion of (x + a)ⁿ are  respectively  84, 280 and 560, find the value of x, a and n.

Solution

It is given that ${\mathrm{T}}_3=84,{ \mathrm{T}}_4=280\text{ and }{\mathrm{T}}_5=560$ .

Now,  ${\mathrm{T}}_3=84\Rightarrow {\mathrm{T}}_3=\mathrm{C}_2^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}–2}{\mathrm{a}}^2=84$_______ (i)

${\mathrm{T}}_4=280\Rightarrow {\mathrm{T}}_4=\mathrm{C}_3^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}–3}{\mathrm{a}}^3=280$________ (ii)

and ${\mathrm{T}}_5=560\Rightarrow {\mathrm{T}}_5=\mathrm{C}_4^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}–4}{\mathrm{a}}^4=560$  ________ (iii)

In order to eliminate x and a, we multiply (i) and (iii) and then divide the product by the square of (ii)

$\therefore \frac{T_3\times T_5}{{\left(T_4\right)}^2}=\frac{84\times 560}{(280{)}^2}=\frac{3}{5}$

$\Rightarrow \frac{T_3}{T_4}\times \frac{T_5}{T_4}=\frac{3}{5}$

$\Rightarrow \frac{3}{n–2}·\frac{x}{a}\times \frac{n–3}{4}·\frac{a}{x}=\frac{3}{5}$   $\left[\because \frac{{\mathrm{T}}_{\mathrm{r}+1}}{{ \mathrm{T}}_{\mathrm{r}}}=\frac{\mathrm{n}–\mathrm{r}+1}{\mathrm{r}}·\frac{\mathrm{a}}{\mathrm{x}}\right]$

$\Rightarrow 5n–15=4n–8$

⇒ n = 7

Now,  $\frac{{\mathrm{T}}_{\mathrm{r}+1}}{{ \mathrm{T}}_{\mathrm{r}}}=\frac{\mathrm{n}–\mathrm{r}+1}{\mathrm{r}}·\frac{\mathrm{a}}{\mathrm{x}}$

$\Rightarrow \frac{{\mathrm{T}}_4}{{ \mathrm{T}}_3}=\frac{\mathrm{n}–2}{3}·\frac{\mathrm{a}}{\mathrm{x}}$

$\Rightarrow \frac{280}{84}=\left(\frac{7–2}{3}\right)\frac{a}{x}$

$\Rightarrow \frac{10}{3}=\frac{5}{3}\left(\frac{a}{x}\right)\Rightarrow x=\frac{a}{2}$

$\therefore {\mathrm{T}}_3=84$

$\Rightarrow C_2^n{x}^{n–2}{a}^2=84$

$\Rightarrow \mathrm{C}_2^7{\mathrm{x}}^5{\mathrm{a}}^2=84 [\because \mathrm{n}=7]$

$\Rightarrow 21{\left(\frac{a}{2}\right)}^5\left(a^2\right)=84 [\because x=a/2]$

$\Rightarrow a^7=2^7\Rightarrow a=2$

$\Rightarrow \mathrm{x}=1 [\because \mathrm{x}=\mathrm{a}/2]$

Hence, x = 1, a = 2 and n = 7.

Illustration -26

If the second, third and fourth term in the expansion of ${\left(x+a\right)}^n$ are 240, 720 and 1080 respectively, then find the value of n.

Solution

Since,

$\begin{array}{l}{\mathrm{t}}_2=240,{\mathrm{t}}_3=720\text{ and }{\mathrm{t}}_4=1080\text{ (given) }\\ C_1^n{x}^{n–1}a=240...\left(1\right)\\ \mathrm{C}_2^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}–2}{\mathrm{a}}^2=720\dots \left(2\right)\\ C_3^n{x}^{n–3}{a}^3=1080.....\left(3\right)\end{array}$

Operate, $\frac{\left(1\right)\left(3\right)}{(2{)}^2}:\frac{C_1^n{x}^{n–1}a·C_3^n{x}^{n–3}{a}^3}{{\left(C_2^n{x}^{n–2}{a}^2\right)}^2}=\frac{240\times 1080}{(720{)}^2}$

$\Rightarrow \left(\frac{\mathrm{C}_1^{\mathrm{n}}}{\mathrm{C}_2^{\mathrm{n}}}\right)\left(\frac{{\mathrm{C}}_3}{\mathrm{C}_2^{\mathrm{n}}}_{\mathrm{n}}\right)=\frac{1}{2}$

$\Rightarrow \left(\frac{2}{n–1}\right)\left(\frac{n–2}{3}\right)=\frac{1}{2}$

Solving, we get n = 5.

8. GREATEST TERM

If $T_r\text{ and }{T}_{r+1}$ be the r^th and (r+1)^th terms in the expansion of ${\left(1+x\right)}^n$, then

$\frac{T_{r+1}}{T_r}=\frac{C_r^n{x}^r}{C_{r–1}^n{x}^{r–1}}=\frac{n–r+1}{r}x$

Let numerically, $T_{r+1}$ be the greatest term in the above expansion. Then $T_{r+1}\ge T_r\text{ or }\frac{T_{r+1}}{T_r}\ge 1$

$\therefore \frac{n–r+1}{r}\left|x\right|\ge 1\text{ or }r\le \frac{(n+1)}{(1+|x\left|\right)}\left|x\right|$……(i)

Now substituting values of n and x in (i), we get  $r\le m+f\text{ or }r\le m$

where m is a positive integer and f is a fraction such that 0 < f < 1.

When n is even $T_{m+1}$ is the greatest term, when n is odd $T_m\text{ and }{T}_{m+1}$ are the greatest terms and both are equal.

Short cut method

To find the greatest term (numerically) in the expansion of ${\left(1+x\right)}^n$.

(i) Calculate m = $\left|\frac{x(n+1)}{x+1}\right|$

(ii) If m is integer, then $T_m$ and $T_{m+1}$ are equal and both are greatest term.

(iii) If m is not integer, there $T_{\left(m\right)+1}$ is the greatest term, where [.] denotes the greatest integral part. 

9. GREATEST COEFFICIENT

(i) If n is even, then greatest coefficient is $C_{n/2}^n$

(ii) If n is odd, then greatest coefficient are $C_{\frac{n+1}{2}}^n\text{ and }C_{\frac{n+3}{2}}^n$

Important Tips

For finding the greatest term in the expansion of ${\left(x+y\right)}^n$. we rewrite the expansion in this form $(x+y{)}^n=x^n{\left[1+\frac{y}{x}\right]}^n$.

Greatest term in (x + y)ⁿ . Greatest term in ${\left(1+\frac{y}{x}\right)}^n$

Illustration -27

Find the greatest term in the expansion of  $(2–3x{)}^9\text{ if }x=\frac{3}{2}$.

Solution

Now, $\frac{t_{r+1}}{t_r}=\left(\frac{n–r+1}{r}\right)\left|\frac{3x}{2}\right|=\left(\frac{10–r}{r}\right)\left|\frac{3x}{2}\right|$

$\Rightarrow \frac{t_{r+1}}{t_r}=\left(\frac{10–r}{r}\right)\left(\frac{3(3/2)}{2}\right)$      $\because x=\frac{3}{2}\left(\text{ given }\right)$

$\therefore \frac{t_{r+1}}{t_r}=\frac{90–9r}{4r}$

Now, for greatest term ${\mathrm{t}}_{\mathrm{r}+1}\ge {\mathrm{t}}_{\mathrm{r}}$  (by definition)

$\Rightarrow 90–9r\ge 4r$

or $r\le \frac{90}{13}=6+\frac{12}{13}\text{ (in the form }I+f)$

Hence, r = 6

$\therefore t_7=C_6^9\left(2{)}^3\right(–3x{)}^6=\frac{9.8.7}{3!}{2}^3{3}^6{\left(\frac{3}{2}\right)}^6=\frac{1}{2}7.3^{13}$

Illustration -28

The Largest term in the expansion of $(3+2x{)}^{50}$; where, $x=\frac{1}{5}$.

Solution

Now, $\frac{t_{t+1}}{t_r}=\left(\frac{n–r+1}{r}\right)\left|\frac{2x}{3}\right|=\left(\frac{51–r}{r}\right)\left|\frac{2x}{3}\right|$

$\Rightarrow \frac{t_{r+1}}{t_r}=\left(\frac{51–r}{r}\right)\left(\frac{2\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}\right)}{3}\right)$    $\because \mathrm{x}=\frac{1}{5}\left(\text{ given }\right)$

$\Rightarrow \frac{t_{r+1}}{t_r}=\left(\frac{102–2r}{15r}\right)$

Now, for greatest term ${\mathrm{t}}_{\mathrm{r}+1}\ge {\mathrm{t}}_{\mathrm{r}}$                 (by definition)

$\Rightarrow 102–2r\ge 15r$

or  $\mathrm{r}\le \frac{102}{17}=6+0\text{ (in the form }\mathrm{I}+f)$

Since, r = 6 an integer

∴   there will be two greatest terms viz. $t_6$ and $t_{6+1}$ terms

${\mathrm{t}}_6=\mathrm{C}_5^{50}\left(3{)}^{45}\right(2\mathrm{x}{)}^5=\mathrm{C}_5^{50}(3{)}^{45}{\left(\frac{2}{5}\right)}^5$  and ${\mathrm{t}}_7=\mathrm{C}_6^{50}\left(3{)}^{50–6}\right(2\mathrm{x}{)}^6=\mathrm{C}_6^{50}(3{)}^{50–6}{\left(\frac{2}{5}\right)}^6$

Illustration -29

If in the expansion of (2x + 5)¹⁰, the greatest term is equal to the middle term, then find the value of x.

Solution

In the expansion of (2x + 5)¹⁰, the middle term = t₆

$\Rightarrow \frac{n+1}{1+\left|\frac{2x}{5}\right|}>5$

$\Rightarrow \frac{11}{5}–1>\left|\frac{2x}{5}\right|$

⇒ |x| < 3

⇒  x ∈ (- 3, 3).

Illustration -30

Find the greatest term in the expansion of $(2–3x{)}^9\text{ if }x=\frac{3}{2}$ .

Solution

Now, $\frac{t_{r+1}}{t_r}=\left(\frac{n–r+1}{r}\right)\left|\frac{3x}{2}\right|=\left(\frac{10–r}{r}\right)\left|\frac{3x}{2}\right|$

$\Rightarrow \frac{t_{r+1}}{t_r}=\left(\frac{10–r}{r}\right)\left(\frac{3(3/2)}{2}\right)$      $\because x=\frac{3}{2}\left(\text{ given }\right)$

$\therefore \frac{t_{r+1}}{t_r}=\frac{90–9r}{4r}$

Now, for greatest term  ${\mathrm{t}}_{\mathrm{r}+1}\ge {\mathrm{t}}_{\mathrm{r}}$   (by definition)

$\Rightarrow 90–9r\ge 4r$

or  $\mathrm{r}\le \frac{90}{13}=6+\frac{12}{13}\text{ (in the form }\mathrm{I}+\mathrm{f})$

Hence, r = 6

∴ ${\mathrm{t}}_7=\mathrm{C}_6^9\left(2{)}^3\right(–3\mathrm{x}{)}^6$ = $= \frac{9·8·7}{3!}{2}^3{3}^6{\left(\frac{3}{2}\right)}^6=\frac{1}{2}7.3^{13}$

Illustration -31

The Largest term in the expansion of $(3+2x{)}^{50}$; where, $x=\frac{1}{5}$.

Solution

Now, $\frac{t_{r+1}}{t_r}=\left(\frac{n–r+1}{r}\right)\left|\frac{2x}{3}\right|=\left(\frac{51–r}{r}\right)\left|\frac{2x}{3}\right|$

$\Rightarrow \frac{t_{r+1}}{t_r}=\left(\frac{51–r}{r}\right)\left(\frac{2(1/5)}{3}\right)$            $\because x=\frac{1}{5}\left(\text{ given }\right)$

$\Rightarrow \frac{t_{r+1}}{t_r}=\left(\frac{102–2r}{15r}\right)$

Now, for greatest term   ${\mathrm{t}}_{\mathrm{r}+1}\ge {\mathrm{t}}_{\mathrm{r}}$             (by definition)

$\Rightarrow 102–2r\ge 15r$

or $\mathrm{r}\le \frac{102}{17}=6+0\text{ (in the form }\mathrm{I}+f)$

Since, r = 6 an integer

∴ there will be two greatest terms viz. ${\mathrm{t}}_6\text{ and }{\mathrm{t}}_{6+1}$ terms

${\mathrm{t}}_6=\mathrm{C}_5^{50}\left(3{)}^{45}\right(2\mathrm{x}{)}^5=\mathrm{C}_5^{50}(3{)}^{45}{\left(\frac{2}{5}\right)}^5$ and ${\mathrm{t}}_7=\mathrm{C}_6^{50}\left(3{)}^{50–6}\right(2\mathrm{x}{)}^6=\mathrm{C}_6^{50}(3{)}^{50–6}{\left(\frac{2}{5}\right)}^6$

Illustration -32

If in the expansion of (2x + 5)¹⁰, the greatest term is equal to the middle term, then find the value of x.

Solution

In the expansion of (2x + 5)¹⁰, the middle term = t₆

$\Rightarrow \frac{\mathrm{n}+1}{1+\left|\frac{2\mathrm{x}}{5}\right|}>5$

$\Rightarrow \frac{11}{5}–1>\left|\frac{2x}{5}\right|$

⇒ |x| < 3

⇒   x ∈ (- 3, 3).

Illustration -33

Find numerically the greatest term in the expansion of ${\left(3–5x\right)}^{11}$ when x = $\frac{1}{5}$.

Solution

Since $(3–5x{)}^{11}=3^{11}{\left(1–\frac{5x}{3}\right)}^{11}$

Now in the expansion of ${\left(1–\frac{5x}{3}\right)}^{11}$, we have

$\frac{{\mathrm{T}}_{\mathrm{r}+1}}{{ \mathrm{T}}_{\mathrm{r}}}=\frac{(11–\mathrm{r}+1)}{\mathrm{r}}\left|–\frac{5\mathrm{x}}{3}\right|$

$=\left(\frac{12–\mathrm{r}}{\mathrm{r}}\right)\left|–\frac{5}{3}\times \frac{1}{5}\right|$     $\left(\because x=\frac{1}{5}\right)=\left(\frac{12–r}{r}\right)\left(\frac{1}{3}\right)=\left(\frac{12–r}{3r}\right)$

$\therefore \frac{{\mathrm{T}}_{\mathrm{r}+1}}{{ \mathrm{T}}_{\mathrm{r}}}\ge 1\Rightarrow \frac{12–\mathrm{r}}{3\mathrm{r}}\ge 1$

$\Rightarrow 4r\le 12$

$\Rightarrow \mathrm{r}\le 3 \therefore \mathrm{r}=2,3$

so, the greatest terms are ${\mathrm{T}}_{2+1}\text{ and }{\mathrm{T}}_{3+1}$.

∴ Greatest terms (where r = 2) = $3^{11}\left|{\mathrm{T}}_{2+1}\right|$

$=3^{11}\left|\mathrm{C}_2^{11}{\left(–\frac{5}{3}\mathrm{x}\right)}^2\right|$

$=3^{11}\left|\mathrm{C}_2^{11}{\left(–\frac{5}{3}\times \frac{1}{5}\right)}^2\right| \left(\because \mathrm{x}=\frac{1}{5}\right)$

$=3^{11}\left|\frac{11.10}{1.2}\times \frac{1}{9}\right|=55\times 3^9$

and greatest term (where r = 3) = $3^{11}\left|{ \mathrm{T}}_{3+1}\right|$

$=3^{11}\left|{}^{11}{\mathrm{C}}_3{\left(–\frac{5}{3}\mathrm{x}\right)}^3\right|$ = $3^{11}\left|\frac{11.10.9}{1.2.3}\times \frac{–1}{27}\right|=55\times 3^9$

From above we say that the values of both greatest terms are equal.

Alternative Method (Short Cut Method) :

Since $(3–5x{)}^{11}=3^{11}{\left(1–\frac{5x}{3}\right)}^{11}=3^{11}{\left(1–\frac{1}{3}\right)}^{11} \left(\because x=\frac{1}{5}\right)$

Now, calculate $\mathrm{m}=\frac{\left|\mathrm{x}\right|(\mathrm{n}+1)}{\left(\right|\mathrm{x}|+1)} \left(–\frac{1}{3}<0\right)$

$=\frac{\left(\left|–\frac{1}{3}\right|\right)(11+1)}{\left(\left|–\frac{1}{3}\right|+1\right)}$= 3

∴ The greatest terms in the expansion are ${\mathrm{T}}_3\text{ and }{\mathrm{T}}_4$

∴ Greatest term (when r = 2) = $3^{11}\left|{ \mathrm{T}}_{2+1}\right|$

$=3^{11}\left|\mathrm{C}_2^{11}{\left(–\frac{1}{3}\right)}^2\right| =3^{11}\left|\frac{11.10}{1.2}\times \frac{1}{9}\right|=55\times 3^9$

and greatest term (when r = 3) = $3^{11}\left|{ \mathrm{T}}_{3+1}\right|$

$=3^{11}\left|\mathrm{C}_3^{11}{\left(–\frac{1}{3}\right)}^3\right|$

$=3^{11}\left|\frac{11.10\times 9}{1.2.3}\times \frac{–1}{27}\right|=55\times 3^9$

From above we say that the values of both greatest terms are equal.

10. PROPERTIES OF BINOMIAL COEFFICIENTS

In the binomial expansion of  $(1+x{)}^n, (1+x{)}^n=C_0^n+C_1^{n}x+C_2^n{x}^2+\dots ..+C_r^n{x}^r+\dots .+C_n^n{x}^n$.

where $C_0^n,C_1^n,C_2^n,\dots \dots ,C_n^n$ are the coefficients of various powers of x and called binomial  coefficients, and they are written as $C_0,C_1,C_2,\dots ..C_n$.

Hence, $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+\dots ..+C_r{x}^r+\dots ..+C_n{x}^n$  …..(i)

(1) The sum of binomial coefficients in the expansion of $(1+x{)}^n\text{ is }{2}^n$.

Putting x=1 in (i), we get $2^n=C_0+C_1+C_2+\dots \dots +C_n$ …..(ii)

(2) Sum of binomial coefficients with alternate signs : Putting x = -1 in (i)

We get,  $0=C_0–C_1+C_2–C_3+\dots \dots$      ………(iii)

(3) Sum of the coefficients of the odd terms in the expansion of ${\left(1+x\right)}^n$ is equal to sum of the  coefficients of even terms and each is equal to $2^{n–1}$.

From (iii), we have $C_0+C_2+C_4+\dots ..=C_1+C_3+C_5+\dots \dots .$             ……(iv)

i.e., sum of coefficients of even and odd terms are equal.

From (ii) and (iv), $C_0+C_2+C_4+\dots ..=C_1+C_3+C_5+\dots ..=2^{n–1}$ ……(v)

(4)  $C_r^n=\frac{n}{r}C_{r–1}^{n–1}=\frac{n}{r}·\frac{n–1}{r–1}C_{r–2}^{n–2}$  and so on.

(5) Sum of product of coefficients : Replacing x by $\frac{1}{x}$ in (i) we get

${\left(1+\frac{1}{x}\right)}^n=C_0+\frac{C_1}{x}+\frac{C_2}{x^2}+\dots \frac{C_n}{x^n}+\dots$(vi)

Multiplying (i) by (vi), we get $\frac{(1+x{)}^{2n}}{x^n}=\left(C_0+C_{1}x+C_2{x}^2+\dots \dots \right)\left(C_0+\frac{C_1}{x}+\frac{C_2}{x^2}+\dots ..\right)$

Now comparing coefficient of $x^r$ on both sides.

We get, $C_{n+r}^{2n}=C_0{C}_r+C_1{C}_{r+1}+\dots \dots C_{n–r}·C_n$  …..(vii)

(6) Sum of squares of coefficients : Putting r=0 in (vii), we get $C_n^{2n}=C_0^2+C_1^2+\dots \dots C_n^2$

(7) $C_r^n+C_{r–1}^n=C_r^{n+1}$

11. USE OF DIFFERENTIATION

This method applied only when the numericals occur as the product of binomial coefficients.

Solution process

(i) If last term of the series leaving the plus or minus sign be m, then divide m by n if q be the quotient and r be the remainder. i.e.,m = nq + r

Then replace x by $x^q$ in the given series and multiplying both sides of expansion by $x^r$.

(ii) After process (i), differentiate both sides, w.r.t. x and put x=1 or -1 or i or – i etc. according to given series.

(iii) If product of two numericals (or square of numericals) or three numericals (or cube of numerical) then differentiate twice or thrice. 

12. USE OF INTEGRATION

This method is applied only when the numericals occur as the denominator of the binomial coefficients.

Solution process

If $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+\dots ..+C_n{x}^n$, then we integrate both sides between the suitable limits which gives the required series.

(i) If the sum contains $C_0,C_1,C_2,\dots \dots .C_n$ with all positive signs, then integrate between limit 0 to 1.

(ii) If the sum contains alternate signs (i.e. +, –) then integrate between limit – 1 to 0.

(iii) If the sum contains odd coefficients i.e., (C₀, C₂, C₄…..) then integrate between –1 to 1.

(iv) If the sum contains even coefficients (i.e., C₁, C₃, C₅….. then subtracting (ii) from (i) and then dividing by 2.

(v) If in denominator of binomial coefficients is product of two numericals then integrate two times, first taking limit between 0 to x and second time take suitable limits.

Illustration -34

Find  the sum of the series $C_0–3C_1+5C_2+\dots +(–1{)}^n(2n+1)C_n$

Solution

We have  $(1–x{)}^n=C_0–C_{1}x+C_2{x}^2–\dots .+C_n(–1{)}^n{x}^n$ ….(i)

Replacing  x by $x^2$ in equation (i), we have

${\left(1–x^2\right)}^n=C_0–C_1{x}^2+C_2{x}^4–\dots +C_n(–1{)}^n{x}^{2n}$….(ii)

Multiplying   throughout   by x, we  have

$x{\left(1–x^2\right)}^n=C_{0}x–C_1{x}^3+C_2{x}^5–\dots +C_n(–1{)}^n{x}^{2n+1}$ ….(iii)

Differentiating  equation (iii) w.r.t. x, we have

${\left(1–x^2\right)}^n–2n{x}^2{\left(1–x^2\right)}^{n–1}=C_0–3C_1{x}^2+5C_2{x}^4–\dots +(–1{)}^n(2n+1)C_n{x}^{2n}$ …(iv)

Putting  x = 1 in equation   (iv), we have

$C_0–3C_1+5C_2–\dots +(–1{)}^n(2n+1)C_n=0$

Illustration -35

Find  the sum  of the series $1^2·{\mathrm{C}}_1+2^2·{\mathrm{C}}_2+3^2·{\mathrm{C}}_3+\dots +{\mathrm{n}}^2·{\mathrm{C}}_{\mathrm{n}}$

Solution

We have $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+\dots .+C_n{x}^n$ ….(i)

Differentiating  equation  (i) w.r.t. x, we have

$n(1+x{)}^{n–1}=C_1+2C_{2}x+3C_3{x}^2+\dots +n{C}_n{x}^{n–1}$ ….(ii)

Multiplying equation (ii) throughout by x, we have

$\mathrm{nx}(1+\mathrm{x}{)}^{\mathrm{n}–1}={\mathrm{C}}_1\mathrm{x}+2{\mathrm{C}}_2{\mathrm{x}}^2+3{\mathrm{C}}_3{\mathrm{x}}^3+\dots +{\mathrm{nC}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$ ….(iii)

Differentiating equation  (iii) w.r.t.  x, we have

$n(1+x{)}^{n–1}+n(n–1\left)\right(1+x{)}^{n–2}$ = ${\mathrm{C}}_1+2^2.{\mathrm{C}}_2\mathrm{x}+3^2.{\mathrm{C}}_3{\mathrm{x}}^2+\dots +{\mathrm{n}}^2.{\mathrm{C}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}–1}$..(iv)

Putting  x = 1 in equation  (iii), we have  

$1^2.C_1+2^2.C_2+3^2.C_3+\dots +n^2.C_n$ = $\mathrm{n}.2^{\mathrm{n}–1}+\mathrm{n}(\mathrm{n}–1), 2^{\mathrm{n}–2}=\left({\mathrm{n}}^2+\mathrm{n}\right)2^{\mathrm{n}–2}=\mathrm{n}(\mathrm{n}+1)2^{\mathrm{n}–2}$

Illustration -36

Find  the sum of  the series $a{C}_0+a^2\frac{C_1}{2}+a^3\frac{C_2}{3}+\dots +a^{n+1}\frac{C_n}{n+1}$

Solution

We have $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+..+C_n{x}^n$            …(i)

Integrating   equation  (i) w.r.t. x, we have

${\int }_0^{\mathrm{a}}(1+\mathrm{x}{)}^{\mathrm{n}}\mathrm{dx}={\int }_0^{\mathrm{a}}\left[{\mathrm{C}}_0+{\mathrm{C}}_1\mathrm{x}+{\mathrm{C}}_2{\mathrm{x}}^2+\dots ..+{\mathrm{C}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\right]\mathrm{dx}$…(ii)

Now, L.H.S. = ${\left[\frac{(1+x{)}^{n+1}}{(n+1)}\right]}_0^a=\frac{(1+a{)}^{n+1}–1}{(n+1)}$

and R.H.S =  ${\left[C_0+C_1\frac{x^2}{2}+C_2\frac{x^3}{3}+..\right]}_0^a=a{C}_0+a^2\frac{C_1}{2}+a^3\frac{C_2}{3}+\dots$

Hence, we have  ${\mathrm{aC}}_0+{\mathrm{a}}^2\frac{{\mathrm{C}}_1}{2}+{\mathrm{a}}^3\frac{{\mathrm{C}}_2}{3}+\dots =\frac{(1+\mathrm{a}{)}^{\mathrm{n}+1}–1}{(\mathrm{n}+1)}$

Illustration -37

Find the  sum of  the series $C_1–\frac{C_2}{2}+\frac{C_3}{3}–\frac{C_4}{4}+\dots +(–1{)}^{n+1}\frac{C_n}{n}$

Solution

We have  $(1–x{)}^n=C_0–C_{1}x+C_2{x}^2–\dots +C_n(–1{)}^n{x}^n$

i.e. $1–(1–x{)}^n=C_{1}x–C_2{x}^2+C_3{x}^3–\dots +C_n(–1{)}^{n+1}{x}^n$

i.e. $\frac{1–(1–\mathrm{x}{)}^{\mathrm{n}}}{\mathrm{x}}={\mathrm{C}}_1–{\mathrm{C}}_2\mathrm{x}+{\mathrm{C}}_3{\mathrm{x}}^2–\dots +{\mathrm{C}}_{\mathrm{n}}(–1{)}^{\mathrm{n}+1}{\mathrm{x}}^{\mathrm{n}–1}$   ….(i)

Integrating equation (i) w.r.t.  x from 0 to 1, we have

${\int }_0^1\frac{1–(1–\mathrm{x}{)}^{\mathrm{n}}}{\mathrm{x}} \mathrm{dx}={\int }_0^1\left[{\mathrm{C}}_1–{\mathrm{C}}_2\mathrm{x}+{\mathrm{C}}_3{\mathrm{x}}^2+\dots \right] \mathrm{dx}$

Now, L.H.S. = ${\int }_0^1\frac{1–{\mathrm{x}}^{\mathrm{n}}}{1–\mathrm{x}}\mathrm{dx}={\int }_0^1\left(1+\mathrm{x}+{\mathrm{x}}^2+\dots +{\mathrm{x}}^{\mathrm{n}–1}\right)\mathrm{dx}$ = ${\left[x+\frac{x^2}{2}+\frac{x^2}{3}+..+\frac{x^n}{n}\right]}_0^1$

$=1+\frac{1}{2}+\frac{1}{3}+\dots .+\frac{1}{n}$

and R.H.S. = ${\left[{\mathrm{C}}_1\mathrm{x}–{\mathrm{C}}_2\frac{{\mathrm{x}}^2}{2}+{\mathrm{C}}_3\frac{{\mathrm{x}}^3}{3}–\dots \right]}_0^1={\mathrm{C}}_1–\frac{{\mathrm{C}}_2}{2}+\frac{{\mathrm{C}}_3}{3}–\dots +(–1{)}^{\mathrm{n}+1}\frac{{\mathrm{C}}_{\mathrm{n}}}{\mathrm{n}+1}$

Hence, we have $C_1–\frac{C_2}{2}+\frac{C_3}{3}+\dots +(–1{)}^{n+1}\frac{C_n}{n}=1+\frac{1}{2}+\frac{1}{3}+\dots ..+\frac{1}{n}$.  

Illustration -38

Find the sum  of the series   $\mathrm{C}_{\mathrm{r}}^{\mathrm{m}}+\mathrm{C}_{\mathrm{r}–1}^{\mathrm{m}}\mathrm{C}_1^{\mathrm{n}}+\mathrm{C}_{\mathrm{r}–2}^{\mathrm{m}}\mathrm{C}_2^{\mathrm{n}}+\dots .+\mathrm{C}_{\mathrm{r}}^{\mathrm{n}}$

where  r < m,  n  and m, n, r ∈ N.

Solution

We have $(1+x{)}^n=C_0^n+C_1^{n}x+C_2^n{x}^2+\dots +C_r^n{x}^r+\dots +C_n^n{x}^n$  ..(i)

and $(1+x{)}^m=C_0^m+C_1^{m}x+\dots +C_{r–2}^m{x}^{r–2}+C_{r–1}^m{x}^{r–1}+C_r^m{x}^r+\dots +C_m^m{x}^m$   ….(ii)

Hence, the  given series = coefficient of $x^r$ in $(1+x{)}^n(1+x{)}^m$ = coefficient of ${\mathrm{x}}^{\mathrm{r}}\text{ in }(1+\mathrm{x}{)}^{\mathrm{m}+\mathrm{n}}{=}^{\mathrm{m}+\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}$.

Illustration -39

Find the sum  of the series ${\mathrm{C}}_1^2+2.{\mathrm{C}}_2^2+3.{\mathrm{C}}_3^2+\dots +\mathrm{n}.{\mathrm{C}}_{\mathrm{n}}^2$

Solution

We have $(1+\mathrm{x}{)}^{\mathrm{n}}={\mathrm{C}}_0+{\mathrm{C}}_1\mathrm{x}+{\mathrm{C}}_2{\mathrm{x}}^2+\dots +\mathrm{n}.{\mathrm{C}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$   …(i)

and ${\left(1+\frac{1}{x}\right)}^n=C_0+C_1\frac{1}{x}+C_2\frac{1}{x^2}+\dots .+C_n\frac{1}{x^n}$   …(ii)

Differentiating   equation  (i) w.r.t. x, we have

$\mathrm{n}(1+\mathrm{x}{)}^{\mathrm{n}–1}={\mathrm{C}}_1+2{\mathrm{C}}_2\mathrm{x}+3{\mathrm{C}}_3{\mathrm{x}}^2+\dots +{\mathrm{nC}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}–1}$

Hence, the given  series  =  coefficient of ${\mathrm{x}}^{–1}\text{ in }{\left(1+\frac{1}{\mathrm{x}}\right)}^{\mathrm{n}}\text{. }\mathrm{n}(1+\mathrm{x}{)}^{\mathrm{n}–1}$

= coefficient of ${\mathrm{x}}^{\mathrm{n}–1}\text{ in }(1+\mathrm{x}{)}^{2\mathrm{n}–1}=\mathrm{n}{.}^{2\mathrm{n}–1}{\mathrm{C}}_{\mathrm{n}–1}$.

13. Multinomial theorem (for positive integral index)

 1. MULTINOMIAL THEOREM

If n is positive integer and $a_1,a_2,a_3,\dots .a_n\in C$ then

${\left(a_1+a_2+a_3+\dots +a_m\right)}^n$ = $\sum \frac{n!}{n_1! n_2! n_3!\dots n_m!}{a}_1^{n_1}{a}_2^{n_2}\dots a_m^{n_m}$

Where $n_1,n_2,n_3,\dots \dots n_m$ are all non-negative integers subject to the condition, .

(1) The coefficient of $a_1^{n_1}·a_2^{n_2}\dots \dots a_m^{n_m}$ in the expansion of ${\left(a_1+a_2+a_3+\dots .a_m\right)}^n$ is $\frac{n!}{n_1! n_2! n_3!\dots n_m!}$

(2) The greatest coefficient in the expansion of ${\left(a_1+a_2+a_3+\dots .a_m\right)}^n$ is $\frac{n!}{(q!{)}^{m–r}[(q+1)!{]}^r}$
Where q is the quotient and r is the remainder when n is divided by m.

(3) If n is +ve integer and $a_1,a_2,\dots ..a_m\in C,a_1^{n_1}.a_2^{n_2}\dots \dots \dots .a_m^{n_m}$ then coefficient of $x^r$ in the expansion of ${\left(a_1+a_{2}x+\dots ..a_m{x}^{m–1}\right)}^n$  is $\sum \frac{n!}{n_1!n_2!n_3!\dots .n_m!}$
Where $n_1,n_2\dots ..n_m$ are all non-negative integers subject to the condition: $n_1+n_2+\dots ..n_m=n$ and $n_2+2n_3+3n_4+\dots .+(m–1)n_m=r$.

(4) The number of distinct or dissimilar terms in the multinomial expansion ${\left(a_1+a_2+a_3+\dots .a_m\right)}^n$ is $C_{m–1}^{n+m–1}$.

2. NUMBER OF TERMS IN THE EXPANSION OF (a + b + c)ⁿ and (a +b + c +d)ⁿ

(a + b + c)ⁿ can be expanded as : (a + b + c)ⁿ  = {(a + b) + c}ⁿ 

= $(a+b{)}^n+C_1^n(a+b{)}^{n–1}(c{)}^1+C_2^n(a+b{)}^{n–2}(c{)}^2+\dots ..+C_n^n{c}^n$

= (n+1) term + n term + (n-1) term + ….. + 1 term

∴ Total number of terms = (n+1) + (n) + (n-1) + ….. + 1 = $\frac{(n+1)(n+2)}{2}$.

Similarly, Number of terms in the expansion of (a + b + c+d)ⁿ  = $\frac{(n+1)(n+2)(n+3)}{6}$.  

Illustration -40

Write the expansion of (3 + x – x²)⁴.

Solution

(3 + x – x²)⁴ = $C_0^4(3+x{)}^4–C_1^4(3+x{)}^3\left(x^2\right)+C_2^4(3+x{)}^2{\left(x^2\right)}^2$$–C_3^4(3+x){\left(x^2\right)}^3+C_4^4{\left(x^2\right)}^4$

= $(3+x{)}^4–4x^2(3+x{)}^3+6x^4(3+x{)}^2–4x^6(3+x)+x^8$

= 81 + 108x + 54x² + 12x³ + x⁴ – 108x² – 108x³ – 36x⁴ – 4x⁵ + 54x⁴ + 36x⁵ + 6x⁶ – 12x⁶ – 4x⁷ + x⁸

= 81 +108x – 54x² – 96x³ + 19x⁴ + 32x⁵ – 6x⁶ – 4x⁷+ x⁸ 

Illustration -41

Find the number of terms in the expansion of (2a + 3b + c)

Solution

No. of terms = $\frac{(n+1)(n+2)}{2}=\frac{(5+1)(5+2)}{2}$= 21

Illustration -42

Find the coefficient of x⁶ in (3+ 2x + x²)⁶.

Solution

(3+ 2x + x²)⁶ = ((3 + 2x) + x²)⁶

$a=C_0^6(3+2x{)}^6.{\left(x^2\right)}^0+C_1^6(3+2x{)}^5.{\left(x^2\right)}^1+$$C_2^6(3+2x{)}^4.{\left(x^2\right)}^2+C_3^6(3+2x{)}^3.{\left(x^2\right)}^3+\dots .$

$=(3+2x{)}^6+6x^2(3+2x{)}^5+15x^4(3+2x{)}^{4}c+20x^6(3+2x{)}^3+\dots .$

$=\left(\sum _{r=0}^{6}C_r^6.3^{6–r}.(2x{)}^r\right)+6x^2\left(\sum _{r=0}^{5}C_r^5.3^{5–r}.(2x{)}^r\right)+$$15x^4\left(\sum _{r=0}^{4}C_r^4.3^{4–r}·(2x{)}^r\right)+20x^6\left(\sum _{r=0}^{3}C_r^3.3^{3–r}.(2x{)}^r\right)+\dots$

The coefficient of x6 in the expansion of (3+ 2x + x²)⁶

=  $\left(C_6^6·3^0{2}^6\right)+6\left(C_4^5·3^1·2^4\right)+15\left(C_2^4·3^2·2^2\right)+20\left(C_0^3·3^3·2^0\right)$

= 64 + 1440 + 3240 + 540 = 5284

Illustration -43

Find the coefficient of $a^3{b}^4{c}^5$ in the expansion of $(\mathrm{bc}+\mathrm{ca}+\mathrm{ab}{)}^6$.

Solution

In this case, $a^3{b}^4{c}^5=\left(ab{)}^x\right(bc{)}^y(ca{)}^z=a^{x+z}.b^{x+y}.c^{y+z}$

z + x = 3, x + y = 4, y + z = 5;  2(x + y + z) = 12; x + y + z = 6. Then x = 1, y = 3, z = 2

Therefore the coefficient of $a^3{b}^4{c}^5$ in the expansion of $(\mathrm{bc}+\mathrm{ca}+\mathrm{ab}{)}^6=\frac{6!}{1! 3! 2!}=60$.

Illustration -44

The coefficient of $x^5$ in the expansion of ${\left(x^2–x–2\right)}^5$.

Solution

General term in the expansion of ${\left({\mathrm{x}}^2–\mathrm{x}–2\right)}^5\text{ is }\sum \frac{5!}{\mathrm{p}! \mathrm{q}! \mathrm{r}!}{\left({\mathrm{x}}^2\right)}^{\mathrm{p}}(–\mathrm{x}{)}^{\mathrm{q}}(–2{)}^{\mathrm{r}}$ is ; where, p, q and r ∈ W

⇒ Coefficient of $x^5$ in the expansion of ${\left(x^2–x–2\right)}^5\text{ is }\sum \frac{5!}{p! q! r!}\left(1{)}^p\right(–1{)}^q(–2{)}^r$

where, p+q+r = 5, 2p+q = 5. The possible value of p, q and r can be

p	0	1	2	3
q	5	3	1	$\xcancel{–1}$
r	0	1	2

∴ the coefficient of  =$x^5=\frac{5!}{1!3!1!}\left(1{)}^1\right(–1{)}^3(–2{)}^1+\frac{5!}{2!1!2!}(1{)}^2(–1{)}^1(–2{)}^2$ + $\frac{5!}{0! 5! 0!}\left(1{)}^0\right(–1{)}^5(–2{)}^0$

40 – 120 – 1 = -81

Illustration -45

Find the total number of terms in the expansion of ${\left(1+a+b\right)}^{10}$ and coefficient of $a^2{b}^3$.

Solution

$(1+a+b{)}^{10}=(1+\overline{a+b}{)}^{10}$ = $\mathrm{C}_0^{\mathrm{n}}\underset{1\text{ tem }}{\underbrace{(\mathrm{a}+\mathrm{b}{)}^0}}+\mathrm{C}_1^{\mathrm{n}}\underset{2\text{ tems }}{\underbrace{(\mathrm{a}+\mathrm{b}{)}^1}}$ + $\mathrm{C}_2^{\mathrm{n}}\underset{3\text{ tems }}{\underbrace{(\mathrm{a}+\mathrm{b}{)}^2}}+\dots +\mathrm{C}_{\mathrm{n}}^{\mathrm{n}}\underset{11\text{ tems }}{\underbrace{(\mathrm{a}+\mathrm{b}{)}^{10}}}$

∴ Total Number of Terms = 1+2+3+….+11 = $\frac{11\times (11+1)}{2}=66$

Also, Total number of terms can be written as ^{10+3 –1}C_3-1 = ¹²C₂ = 66

Secondly, Coefficient  of  a²b³ = $\frac{10!}{2!\times 3!\times 15!}$ =  2520

Illustration -46

If the number of terms in the expansion of $(\alpha +2\beta +3\gamma +\delta {)}^n$ are 56, then find the value of n.

Solution

Since, the number of terms in the expansion of $(\alpha +2\beta +3\gamma +\delta {)}^n$ are $C_3^{n+3}$

$\therefore \frac{(n+3)(n+2)(n+1)}{3!}=56$

⇒ $\boxed{n=5}$

14. Binomial theorem for any Index

Statement

$(1+x{)}^n=1+nx+\frac{n(n–1)x^2}{2!}+\frac{n(n–1)(n–2)}{3!}{x}^3$$+\dots .+\frac{n(n–1)\dots \dots (n–r+1)}{r!}{x}^r+\dots$term up to $\infty$

When n is a negative integer or a fraction, where -1 < x < 1, otherwise expansion will not be possible.

If x<1, the terms of the above expansion go on decreasing and if x be very small a stage may be reached when we may neglect the terms containing higher power of x in the expansion, then $(1+x{)}^n=1+nx$.

Important Tips

Expansion is valid only when -1 < x < 1.

ⁿC_r cannot be used because it is defined only for natural number, so ⁿC_r will be written as

$\frac{\left(n\right)(n–1)\dots \dots (n–r+1)}{r!}$

The number of terms in the series is infinite.

If first term is not 1, then make first term unity in the following way, $(\mathrm{x}+\mathrm{y}{)}^{\mathrm{n}}={\mathrm{x}}^{\mathrm{n}}{\left[1+\frac{\mathrm{y}}{\mathrm{x}}\right]}^{\mathrm{n}},\text{ if }\left|\frac{\mathrm{y}}{\mathrm{x}}\right|<1$.

General term : $T_{r+1}=\frac{n(n–1)(n–2)\dots \dots (n–r+1)}{r!}{x}^r$

15. SOME IMPORTANT EXPANSIONS

(i) $(1+x{)}^n=1+nx+\frac{n(n–1)}{2!}{x}^2+\dots \dots ..+$$\frac{n(n–1)(n–2)\dots \dots (n–r+1)}{r!}{x}^r+\dots \dots$

(ii) $(1–x{)}^n=1–nx+\frac{n(n–1)}{2!}{x}^2–\dots \dots .+$$\frac{n(n–1)(n–2)\dots ..(n–r+1)}{r!}(–x{)}^r+\dots \dots$

(iii) $(1–x{)}^{–n}=1+nx+\frac{n(n+1)}{2!}{x}^2+\frac{n(n+1)(n+2)}{3!}{x}^3+\dots ..+$$\frac{n(n+1)\dots \dots (n+r–1)}{r!}{x}^r+\dots ..$

(iv) $(1+x{)}^{–n}=1–nx+\frac{n(n+1)}{2!}{x}^2–\frac{n(n+1)(n+2)}{3!}{x}^3+\dots ..+$$\frac{n(n+1)\dots \dots (n+r–1)}{r!}(–x{)}^r+\dots \dots$

(a) Replace n by 1 in (iii) : $(1–x{)}^{–1}=1+x+x^2+\dots ..+x^r+\dots \dots \infty$,  General term, $T_{r+1}=x^r$

(b) Replace n by 1 in (iv) : $(1+x{)}^{–1}=1–x+x^2–x^3+\dots ..+(–x{)}^r+\dots \dots \infty$,  General term, $T_{r+1}=(–x{)}^r$.

(c) Replace n by 2 in (iii) : $(1–x{)}^{–2}=1+2bx+3x^2+\dots ..+(r+1)x^r+\dots ..\infty$,  General term, $T_{r+1}=(r+1)x^r$.

(d) Replace n by 2 in (iv) : $(1+x{)}^{–2}=1–2x+3x^2–4x^3+\dots \dots +(r+1\left)\right(–x{)}^r+\dots ..\infty$, General term, $T_{r+1}=(r+1)(–x{)}^r$.

(e) Replace n by 3 in (iii) : $(1–x{)}^{–3}=1+3x+6x^2+10x^3+\dots ..+$$\frac{(r+1)(r+2)}{2!}{x}^r+\dots \dots \dots .\infty$, General term, $T_{r+1}=(r+1)(r+2)/2!.x^r$

(f) Replace n by 3 in (iv) : $(1+x{)}^{–3}=1–3x+6x^2–10x^3+\dots ..+$$\frac{(r+1)(r+2)}{2!}(–x{)}^r+\dots ..\infty$, General term, $T_{r+1}=\frac{(r+1)(r+2)}{2!}(–x{)}^r$

Illustration 47

Find the range of values of x for which the expansion of $\frac{1}{\sqrt{\left(3+2x\right)}}$ can be expanded.

Solution

Now, $\frac{1}{\sqrt{3+2x}}=(3+2x{)}^{–\frac{1}{2}}={\left\{3\left(1+\frac{2x}{3}\right)\right\}}^{–\frac{1}{2}}=\frac{1}{\sqrt{3}}{\left(1+\frac{2x}{3}\right)}^{–\frac{1}{2}}$

$\therefore {\left(1+\frac{2x}{3}\right)}^{\frac{1}{2}}$ can be expanded if and only if $\left|\frac{2x}{3}\right|<1$. Hence,  $x\in \left(–\frac{3}{2},\frac{3}{2}\right)$

Illustration 48

Find the coefficient of  in the expansion of ${\left(1–2x\right)}^{\raisebox{1ex}{$–1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ .

Solution

The given expansion ${\left(1–2x\right)}^{\raisebox{1ex}{$–1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is defined iff $x\in \left(–\frac{1}{2},\frac{1}{2}\right)$

Now, Coefficient of $x^r$ in the expansion of ${\left(1–2x\right)}^{\raisebox{1ex}{$–1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is

$\frac{\left(–\frac{1}{2}\right)\left(–\frac{1}{2}–1\right)\left(–\frac{1}{2}–2\right)\dots \left(–\frac{1}{2}–(r–1)\right)}{r!}(–2{)}^r$

or  $(–1{)}^r\frac{1}{2^r}\frac{1·3·5\dots (2r–1)}{r!}(–1{)}^r{2}^r=\frac{1·3·5\dots (2r–1)}{r!}$

Illustration 49

If x is so small that $x^2$ and higher powers of x can be neglected, then find the value of $\frac{\sqrt{1+x}+\sqrt[3]{(1–x{)}^2}}{1+x+\sqrt{1+x}}$

Solution

Let E = $\frac{\sqrt{1+x}+\sqrt[3]{(1–x{)}^2}}{1+x+\sqrt{1+x}}=\frac{(1+x{)}^{\frac{1}{2}}+(1–x{)}^{\frac{2}{3}}}{(1+x)+(1+x{)}^{\frac{1}{2}}}$

Since, given that square and higher powers of x can be neglected

$\Rightarrow \mathrm{E}=\frac{\left(1+\frac{1}{2}\mathrm{x}\right)+\left(1–\frac{2}{3}\mathrm{x}\right)}{(1+\mathrm{x})+\left(1+\frac{1}{2}\mathrm{x}\right)}=\frac{2–\frac{1}{6}\mathrm{x}}{2+\frac{3}{2}\mathrm{x}}=\frac{\bcancel{2}\left(1–\frac{1}{12}\mathrm{x}\right)}{\bcancel{2}\left(1+\frac{3}{4}\mathrm{x}\right)}$

Again neglecting square and higher powers of x, we have

$E=\left(1–\frac{1}{12}x\right){\left\{\left(1+\frac{3}{4}x\right)\right\}}^{–1}$ = $\left(1–\frac{1}{12}x\right)\left\{1+(–1)\left(\frac{3}{4}x\right)\right\}$ = $\left(1–\frac{1}{12}x\right)\left(1–\frac{3}{4}x\right)$

$\therefore E=1–\frac{1}{12}x–\frac{3}{4}x+\underset{\mathrm{neglecting}}{\frac{1}{16}{x}^2}=1–\frac{5}{6}x$

Illustration 50

Find the range of x for which the binomial expansions of $(3–4x{)}^{–5/2}$ is valid

Solution

$(3–4x{)}^{–5/2}=3^{–5/2}{\left(1–\frac{4x}{3}\right)}^{–5/2}$

This expansion is valid  

$\Rightarrow \left|\frac{4x}{3}\right|<1\Rightarrow \left|x\right|<\frac{3}{4}\Rightarrow –\frac{3}{4}<x<\frac{3}{4}$

The range of x is $\left(–\frac{3}{4},\frac{3}{4}\right)$ 

Illustration 51

Write down the first three terms in the expansion of ${(4 + 9x)}^{–7}$

Solution

$(4+9x{)}^{–7}=4^{–7}{\left(1+\frac{9x}{4}\right)}^{–7}$

$=4^{–7}\left[1–C_1^7\left(\frac{9x}{4}\right)+C_2^8{\left(\frac{9x}{4}\right)}^2\dots \right]$

$=4^{–7}\left[1–\frac{63x}{4}+\frac{567x^2}{4}+\dots \right]$

∴ The first three terms are  $4^{–7},–63\times 4^{–8}x, 567\times 4^{–8}{x}^2$

Illustration 52

Write the general term in the expansion of ${\left(3–\frac{5x}{4}\right)}^{–1/2}$

Solution

${\left(3–\frac{5x}{4}\right)}^{–1/2}=3^{–1/2}{\left(1–\frac{5x}{12}\right)}^{–1/2}$

$T_{r+1}=3^{–1/2}\frac{\left(–\frac{1}{2}\right)\left(–\frac{1}{2}–1\right)\left(–\frac{1}{2}–2\right)\dots \left(–\frac{1}{2}–r+1\right)}{r!}{\left(–\frac{5x}{12}\right)}^r$

$=3^{–1/2}\frac{(–1{)}^{2r}}{r!}\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\left(\frac{5}{2}\right)\dots \left(\frac{2r–1}{2}\right){\left(\frac{5x}{12}\right)}^r$

$=3^{–1/2}\times \frac{1\times 3\times 5\times \dots \times (2r–1)}{r!}{\left(\frac{5x}{24}\right)}^r$

Illustration 53

If |x| is so small that $x^4$ and higher powers of x may be neglected, then find an approximate value of $\sqrt[4]{x^2+81}–\sqrt[4]{x^2+16}$.

Solution

$\sqrt[4]{x^2+81}–\sqrt[4]{x^2+16}={\left(81+x^2\right)}^{1/4}–{\left(16+x^2\right)}^{1/4}$

$={81}^{1/4}{\left(1+\frac{x^2}{81}\right)}^{1/4}–{16}^{1/4}\left(1+\frac{x^2}{16}\right)$

$=3\left(1+\frac{1}{4}·\frac{x^2}{81}\right)–2·\left(1+\frac{1}{4}·\frac{x^2}{16}\right)=1–\frac{19}{864}{x}^2$

Illustration 54

Find the value of $\sqrt[5]{32.16}$ corrected to 4 decimal places

Solution

$\sqrt[5]{32.16}=(32+0.16{)}^{1/5}$

$=(32{)}^{1/5}{\left(1+\frac{0.16}{32}\right)}^{1/5}=2[1+0.005{]}^{1/5}$

$=2\left[1+\frac{1}{5}(0.005)+\frac{\frac{1}{5}\left(–\frac{4}{5}\right)}{2!}(0.005{)}^2+\dots \right]$

$=2\left[1+0.001–\frac{2}{25}(0.000025)+\dots .\right]$

= 2(1.001) = 2.002

$\therefore \sqrt[5]{32.16}=2.002$

16. Some useful results FOR DIVISIBILITY problems

Let (1 + x)ⁿ = 1 + ⁿC₁x + ⁿC₂x² + … + ⁿC_nxⁿ.

In any divisibility problem, we have to identify x and n. The number by which division is to be made can be x, x² or x³, but the number in the base is always expressed in form of (1 + x).

Illustration 55

Find the remainder when 7¹⁰³ is divided by 25.

Solution

7¹⁰³ = 7(50 – 1)⁵¹ = 7(50⁵¹ – ⁵¹C₁ 50⁵⁰ + ⁵¹C₂ 50⁴⁹ – … – 1)

7¹⁰³ = 7(50⁵¹ – ⁵¹C₁ 50⁵⁰ + … + ⁵¹C₅₀ 50) – 7 – 18 + 18

7¹⁰³ = 7(50⁵¹ – ⁵¹C₁ 50⁵⁰ + … + ⁵¹C₅₀ 50) – 25 + 18

⇒   remainder is 18.   

17. Expansion with Three or Four consecutive terms or Coefficients

(1) If consecutive coefficients are given: In this case divide consecutive coefficients pair wise. We get equations and then solve them.

(2) If consecutive terms are given : In this case divide consecutive terms pair wise i.e. if four consecutive terms be $T_r,T_{r+1},T_{r+2},T_{r+3}$ then find $\frac{T_r}{T_{r+1}},\frac{T_{r+1}}{T_{r+2}},\frac{T_{r+2}}{T_{r+3}}\Rightarrow {\lambda }_1,{\lambda }_2,{\lambda }_3$ (say) then divide ${\lambda }_1$ by ${\lambda }_2$ and ${\lambda }_3$ by  and solve.

(3) Method for finding terms free from radical or rational terms in the expansion of ${(a^{1/P}+b^{1/q})}^{\mathbf{N}}\mathbf{\forall }\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }$ prime numbers :

Find the general term $T_{r+1}=C_r^N{\left(a^{1/p}\right)}^{N–r}{\left(b^{1/q}\right)}^r$ =  $C_r^N a^{\frac{N–r}{p}}·b^{\frac{r}{q}}$

Putting the values of $0\le r\le N$, when indices of a and b are integers.

Note : Number of irrational terms = Total terms – Number of rational terms.

Illustration 56

If $a_1,a_2,a_3,a_4$ are the coefficients of any four consecutive terms in the expansion of $(1+x{)}^n$, then $\frac{a_1}{a_1+a_2}+\frac{a_3}{a_3+a_4}$ =

(A) $\frac{a_2}{a_2+a_3}$        (B) $\frac{1}{2}\frac{a_2}{a_2+a_3}$       (C) $\frac{2a_2}{a_2+a_3}$          (D) $\frac{2a_3}{a_2+a_3}$

Solution

(C)

Let $a_1,a_2,a_3,a_4$ be respectively the coefficients of $(r+1{)}^{\text{th }},(r+2{)}^{\text{th }},(r+3{)}^{\text{th }},(r+4{)}^{th}$ terms in the expansion of ${\left(1+x\right)}^n$.

Then $a_1=C_r^n, a_2=C_{r+1}^n, a_3=C_{r+2}^n, a_4=C_{r+3}^n$.

Now,  $\frac{a_1}{a_1+a_2}+\frac{a_3}{a_3+a_4}=\frac{C_r^n}{C_r^n+C_{r+1}^n}+\frac{C_{r+2}^n}{C_{r+2}^n+C_{r+3}^n}$ = $\frac{C_r^n}{C_{r+1}^{n+1}}+\frac{C_{r+2}^n}{C_{r+3}^{n+1}}=\frac{C_r^n}{\frac{n+1}{r+1}C_r^n}+\frac{C_{r+2}^n}{\frac{n+1}{r+3}C_{r+2}^n}$

$=\frac{r+1}{n+1}+\frac{r+3}{n+1}=\frac{2(r+2)}{n+1}$ = $= 2.\frac{C_{r+1}^n}{C_{r+2}^{n+1}}=2·\frac{C_{r+1}^n}{C_{r+2}^{n+1}}=2·\frac{C_{r+1}^n}{C_{r+1}^n+C_{r+2}^n}=\frac{2a_2}{a_2+a_3}$

Illustration 57

The coefficient of two consecutive terms in the expansion of ${\left(1+x\right)}^n$ will be equal, if

(A) n is any integer             (B) n is an odd integer

(C) n is an even integer      (D) None of these

Solution

(B)

Let  consecutive terms are and $C_r^n\text{ and }C_{r+1}^n$

$\Rightarrow \frac{n!}{(n–r)! r!}=\frac{n!}{(n–r–1)! (r+1)!}$

$\Rightarrow \frac{1}{(n–r)(n–r–1)! r!}=\frac{1}{(n–r–1)! (r+1)r!}$

$\Rightarrow r+1=n–r\Rightarrow n=2r+1$. Hence n is odd

Illustration 58

If in the expansion of ${\left(1+x\right)}^n$, a, b, c are three consecutive coefficients, then n=

(A) $\frac{ac+ab+bc}{b^2+ac}$     (B) $\frac{2ac+ab+bc}{b^2–ac}$     (C) $\frac{ab+ac}{b^2–ac}$       (D) None of these

Solution

(B)

Trick : Here $a=C_r^n, b=C_{r+1}^n$ and $c=C_{r+2}^n$

Put n=2, r=0, then option (b) holds the condition i.e. $n=\frac{2ac+ab+bc}{b^2–ac}$

Illustration 59

The number of integral terms in the expansion of  $(\sqrt{3}+\sqrt[8]{5}{)}^{256}$ is

(A) 32                   (B) 33                       (C) 34                          (D) 35

Solution

(B)

$T_{r+1}=C_r^{256} . 3^{\frac{256–r}{2}}. 5^{\frac{r}{8}}$

First term = $\mathrm{C}_0^{256} 3^{128} 5^0$ = integer and after eight terms, i.e., 9^th term = $C_8^{256} 3^{124}.5^1$ = integer

Continuing like this, we get an A.P., $1^{\text{st }},9^{\text{th }}\dots \dots .{257}^{\text{th }}$; $T_n=a+(n–1)d\Rightarrow 257=1+(n–1)8\Rightarrow n=33$ 

Illustration 60

The number of irrational terms in the expansion of $(\sqrt{3}+\sqrt[8]{5}{)}^{256}$ is

(A) 97                   (B) 98                       (C) 96                          (D) 99

Solution

(A)

$T_{r+1}=C_r^{100} 5^{\frac{100–r}{8}}·2^{\frac{r}{6}}$

As 2 and 5 are co-prime. $T_{r+1}$ will be rational if 100-r is multiple of 8 and r is multiple of 6 also

$0\le r\le 100$

∴ r = 0,6,12……96 ; ∴ 100 – r = 4,10,16…..100     ……(i)

But  is to be multiple of 8.

So, = 100-r = 0, 8, 16, 24,……96          …..(ii)

Common terms in (i) and (ii) are 16, 40, 64, 88.

∴ r = 84, 60, 36, 12 give rational terms

∴ The number of irrational terms  = 101 – 4 = 97. 

18. Miscellaneous concepts and problems

PROBLEMS INVOLVING THE GREATEST INTEGER FUNCTION

If $(\sqrt{A}+B{)}^n=I+f$ where I and n are positive integers, n being odd and $0\le f<1$ then $(I+f).f=K^n$

where  =$A–B^2=K>0\text{ and }\sqrt{A}–B<1$ .

Note :  If n is even integer then $(\sqrt{A}+B{)}^n+(\sqrt{A}–B{)}^n=I+f+f‘$

Hence L.H.S. and I are integers.

∴ f + f’ is also integer;  $\Rightarrow f+f‘=1; \therefore f‘=(1–f)$

Hence $(\mathrm{I}+\mathrm{f})(1–\mathrm{f})=(\mathrm{I}+\mathrm{f})\mathrm{f}‘$ = $(\sqrt{\mathrm{A}}+\mathrm{B}{)}^{\mathrm{n}}(\sqrt{\mathrm{A}}–\mathrm{B}{)}^{\mathrm{n}}={\left(\mathrm{A}–{\mathrm{B}}^2\right)}^{\mathrm{n}}={\mathrm{K}}^{\mathrm{n}}$.

Illustration 61

Let $R=(5\sqrt{5}+11{)}^{2n+1}$ and $f=R–\left[R\right]$, where [.] denotes the greatest integer function. The value of R.f is

(A) $4^{2n+1}$       (B) $4^{2n}$              (C) $4^{2n–1}$        (D) $4^{–2n}$

Solution

(A) Since $(5\sqrt{5}–11)(5\sqrt{5}+11)=4$

$\Rightarrow 5\sqrt{5}–11=\frac{4}{5\sqrt{5}+11}$,

$\because 0<5\sqrt{5}–11<1=0<(5\sqrt{5}–11{)}^{2n+1}<1$,

for positive integral n.

Again, $(5\sqrt{5}+11{)}^{2n+1}–(5\sqrt{5}–11{)}^{2n+1}$

$=2\left\{{}^{2n+1}{C}_1(5\sqrt{5}{)}^{2n}·11+{}^{2n+1}C_3(5\sqrt{5}{)}^{2n–2}\right.$$\left.\times {11}^3+\dots .+{}^{2n+1}C_{2n+1}{11}^{2n+1}\right\}$

$=2\left\{{}^{2n+1}C_1(125{)}^n·11+{}^{2n+1}C_3(125{)}^{n–1}{11}^3+.\right.$$\left.\dots +{}^{2n+1}C_{2n+1}{11}^{2n+1}\right\}$

= 2k, (for some positive integer k)

Let $f^{‘}=(5\sqrt{5}–11{)}^{2n+1},\text{ then }[R]+f–f^{‘}=2k$,

$\Rightarrow f–f^{‘}=2k–\left[R\right]\Rightarrow f–f^{‘}$ is an integer.

But, $0\le f<1;0<f^{‘}<1\Rightarrow –1<f–f^{‘}<1$

$\Rightarrow f–f^{‘}=0$ (integer) $\Rightarrow f=f^{‘}$

$\because Rf=R{f}^{‘}=(5\sqrt{5}+11{)}^{2n+1}(5\sqrt{5}–11{)}^{2n+1}$

$={\left[(5\sqrt{5}{)}^2–{11}^2\right]}^{2n+1}=4^{2n+1}$.

Illustration 62

The greatest integer less than or equal to $(\sqrt{2}+1{)}^6$ is

(A) 196                 (B) 197                 (C)198             (D) 199

Solution

(B)

Let $(\sqrt{2}+1{)}^6=k+f$, where k is integral part and f the fraction $(0\le f<1)$.

Let $(\sqrt{2}–1{)}^6=f^{‘}, \left(0<f^{‘}<1\right)$,

Since $0<(\sqrt{2}–1)<1$

Now, $k+f+f^{‘}=(\sqrt{2}+1{)}^6+(\sqrt{2}–1{)}^6$

$=2\left\{{}^{6}C_0·2^3+{}^{6}C_2·2^2+{}^{6}C_4·2+{}^{6}C_6\right\}=198$…..(i)

$f+f^{‘}=198–k=$an integer

But, $0\le f<1\text{ and }0<f^{‘}<1\Rightarrow 0<\left(f+f^{‘}\right)<2$,

$\Rightarrow f+f^{‘}=1$,      ($\because$ f+f’ is an integer)

∴ By (i), I =198 – (f+f’) = 198  – 1 = 197.