NCERT Solutions for Exercise 8.2 Class 11 Maths Chapter 8 - Binomial Theorem

Binomial Theorem Class 11 Chapter 8 Exercise: 8.1

Question:1 Find the coefficient of

$x^5$ in $(x+3{)}^8$

Answer:

As we know that the $(r+1{)}^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b{)}^n$ is given by

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

Now let's assume $x^5$ happens in the $(r+1{)}^{th}$ term of the binomial expansion of $(x+3{)}^8$

So,

$T_{r+1}{=}^8{C}_r{x}^{8-r}{3}^r$

On comparing the indices of x we get,

$r=3$

Hence the coefficient of the $x^5$ in $(x+3{)}^8$ is

${}^8{C}_3\times 3^3=\frac{8!}{5!3!}\times 9=\frac{8\times 7\times 6}{3\times 2}\times 9=1512$

Question:2 Find the coefficient of$a^5{b}^7$ in $(a-2b{)}^{12}$

Answer:

As we know that the $(r+1{)}^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b{)}^n$ is given by

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

Now let's assume$a^5{b}^7$ happens in the $(r+1{)}^{th}$ term of the binomial expansion of $(a-2b{)}^{12}$

So,

$T_{r+1}{=}^{12}{C}_r{x}^{12-r}(-2b{)}^r$

On comparing the indices of x we get,

$r=7$

Hence the coefficient of the $a^5{b}^7$ in $(a-2b{)}^{12}$ is

$$\begin{gathered} {\Rightarrow }^{12}{C}_7\times (-2{)}^7=\frac{12!}{5!7!}\times (-128) \\ =\frac{12\times 11\times 10\times 9\times 8}{5\times 4\times 3\times 2}\times (-128) \\ =-(729)(128) \\ =-101376 \end{gathered}$$

Question:3 Write the general term in the expansion of

$(x^2-y{)}^6$

Answer:

As we know that the general $(r+1{)}^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b{)}^n$ is given by

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

So the general term of the expansion of $(x^2-y{)}^6$ :

$T_{r+1}{=}^6{C}_r(x^2{)}^{6-r}(-y{)}^r=(-1{)}^r{\times }^6{C}_r{x}^{12-2r}{y}^r$.

Question:4 Write the general term in the expansion of

$(x^2-xy{)}^{12},x\ne 0$

Answer:

As we know that the general $(r+1{)}^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b{)}^n$ is given by

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

So the general term of the expansion of $(x^2-xy{)}^{12},$is

$$\begin{gathered} \Rightarrow T_{r+1} \\ {=}^{12}{C}_r(x^2{)}^{12-r}(-xy{)}^r \\ =(-1{)}^r{\times }^{12}{C}_r{x}^{24-2r+r}{y}^r \\ =(-1{)}^r{\times }^{12}{C}_r{x}^{24-r}{y}^r \end{gathered}$$.

Question:5 Find the 4^th term in the expansion of $(x-2y{)}^{12}$.

Answer:

As we know that the general $(r+1{)}^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b{)}^n$ is given by

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

So the $4^{th}$ term of the expansion of $(x-2y{)}^{12}$ is

$$\begin{gathered} \Rightarrow T_4=T_{3+1} \\ {=}^{12}{C}_3(x{)}^{12-3}(-2y{)}^3 \\ =(-2{)}^3{\times }^{12}{C}_3{x}^9{y}^3 \\ =-8\times \frac{12!}{3!9!}{x}^9{y}^3 \end{gathered}$$

$$\begin{gathered} =-8\times \frac{12\times 11\times 10}{3\times 2}\times x^9{y}^3 \\ =-8\times 220\times x^9{y}^3 \\ =-1760x^9{y}^3 \end{gathered}$$.

Question:6 Find the 13^th term in the expansion of ${(9x-\frac{1}{3\sqrt{x}})}^{18},x\ne 0$

Answer:

As we know that the general $(r+1{)}^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b{)}^n$ is given by

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

So the ${13}^{th}$ term of the expansion of ${(9x-\frac{1}{3\sqrt{x}})}^{18}$ is

$$\begin{gathered} \Rightarrow T_{13}=T_{12+1} \\ {=}^{18}{C}_{12}(9x{)}^{18-12}{\left(\frac{1}{3\sqrt{x}}\right)}^{12} \\ =\frac{18!}{12!6!}\times 9^6{\left(\frac{1}{3}\right)}^{12}\times x^{6-6} \end{gathered}$$

$=\frac{18\times 17\times 16\times 15\times 14\times 13}{6\times 5\times 4\times 3\times 2}\times 9^6\left(\frac{1}{3^{12}}\right)$

$=18564$

Question:7 Find the middle terms in the expansion of${(3-\frac{x^3}{6})}^7$

Answer:

As we know that the middle terms in the expansion of $(a+b{)}^n$ when n is odd are,

${\left(\frac{n+1}{2}\right)}^{th}termand{(\frac{n+1}{2}+1)}^{th}term$

Hence the middle term of the expansion ${(3-\frac{x^3}{6})}^7$ are

${\left(\frac{7+1}{2}\right)}^{th}termand{(\frac{7+1}{2}+1)}^{th}term$

Which are $4^{th}termand{5}^{th}term$

Now,

As we know that the general $(r+1{)}^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b{)}^n$ is given by

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

So the $4^{th}$ term of the expansion of ${(3-\frac{x^3}{6})}^7$is

$$\begin{gathered} \Rightarrow T_4=T_{3+1} \\ {=}^7{C}_3(3{)}^{7-3}{(-\frac{x^3}{6})}^3 \\ ={(-\frac{1}{6})}^3\times 3^4{\times }^7{C}_3\times x^9 \\ ={(-\frac{1}{6})}^3\times 3^4\times \frac{7!}{3!4!}\times x^9 \\ =-\frac{3\times 3\times 3\times 3}{6\times 6\times 6}\times \frac{7\times 6\times 5}{3\times 2}\times x^9 \end{gathered}$$

$=-\frac{105}{8}{x}^9$

And the $5^{th}$ Term of the expansion of ${(3-\frac{x^3}{6})}^7$is,

$$\begin{gathered} \Rightarrow T_5=T_{4+1} \\ {=}^7{C}_4(3{)}^{7-4}{(-\frac{x^3}{6})}^4 \\ ={(-\frac{1}{6})}^4\times 3^3{\times }^7{C}_4\times x^{12} \\ ={(-\frac{1}{6})}^4\times 3^3\times \frac{7!}{3!4!}\times x^{12} \\ =\frac{3\times 3\times 3}{6\times 6\times 6\times 6}\times \frac{7\times 6\times 5}{3\times 2}\times x^{12} \end{gathered}$$

$=\frac{35}{48}{x}^{12}$

Hence the middle terms of the expansion of given expression are

$-\frac{105}{8}{x}^{9}and\frac{35}{48}{x}^{12}.$

Question:8 Find the middle terms in the expansion of${(\frac{x}{3}+9y)}^{10}$

Answer:

As we know that the middle term in the expansion of $(a+b{)}^n$ when n is even is,

${(\frac{n}{2}+1)}^{th}term$,

Hence the middle term of the expansion ${(\frac{x}{3}+9y)}^{10}$ is,

${(\frac{10}{2}+1)}^{th}term$

Which is $6^{th}term$

Now,

As we know that the general $(r+1{)}^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b{)}^n$ is given by

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

So the $6^{th}$ term of the expansion of ${(\frac{x}{3}+9y)}^{10}$is

$$\begin{gathered} \Rightarrow T_6=T_{5+1} \\ {=}^{10}{C}_5{\left(\frac{x}{3}\right)}^{10-5}{\left(9y\right)}^5 \end{gathered}$$

$={\left(\frac{1}{3}\right)}^5\times 9^5{\times }^{10}{C}_5\times x^5{y}^5$

$={\left(\frac{1}{3}\right)}^5\times 9^5\times \left(\frac{10!}{5!5!}\right)\times x^5{y}^5$

$=\left(\frac{1}{3^5}\right)\times 9^5\times \left(\frac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2}\right)\times x^5{y}^5$

$=61236x^5{y}^5$

Hence the middle term of the expansion of ${(\frac{x}{3}+9y)}^{10}$ is nbsp; $61236x^5{y}^5$.

Question:9 In the expansion of $(1+a{)}^{m+n}$ , prove that coefficients of $a^m$ and $a^n$ are equal

Answer:

As we know that the general $(r+1{)}^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b{)}^n$ is given by

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

So, the general $(r+1{)}^{th}$ term $T_{r+1}$ in the binomial expansion of $(1+a{)}^{m+n}$ is given by

$T_{r+1}{=}^{m+n}{C}_r{1}^{m+n-r}{a}^r{=}^{m+n}{C}_r{a}^r$

Now, as we can see $a^m$ will come when $r=m$ and $a^n$ will come when $r=n$

So,

Coefficient of $a^m$ :

$K_{a^m}{=}^{m+n}{C}_m=\frac{(m+n)!}{m!n!}$

CoeficientCoefficient of $a^n$ :

$K_{a^n}{=}^{m+n}{C}_n=\frac{(m+n)!}{m!n!}$

As we can see $K_{a^m}=K_{a^n}$.

Hence it is proved that the coefficients of $a^m$ and $a^n$ are equal.

Question:10 The coefficients of the $(r-1)$^th , r^th and $(r+1)$^th terms in the expansion of $(x+1{)}^n$ are in the ratio 1 : 3 : 5. Find n and r.

Answer:

As we know that the general $(r+1{)}^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b{)}^n$ is given by

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

So,

$(r+1{)}^{th}$ Term in the expansion of $(x+1{)}^n$:

$T_{r+1}{=}^n{C}_r{x}^{n-r}{1}^r{=}^n{C}_r{x}^{n-r}$

$r^{th}$ Term in the expansion of $(x+1{)}^n$:

$T_r{=}^n{C}_{r-1}{x}^{n-r+1}{1}^{r-1}{=}^n{C}_{r-1}{x}^{n-r+1}$

$(r-1{)}^{th}$ Term in the expansion of $(x+1{)}^n$:

$T_{r-1}{=}^n{C}_{r-2}{x}^{n-r+2}{1}^{r-2}{=}^n{C}_{r-2}{x}^{n-r+2}$

Now, As given in the question,

$T_{r-1}:T_r:T_{r+1}=1:3:5$

${}^n{C}_{r-2}{:}^n{C}_{r-1}{:}^n{C}_r=1:3:5$

$\frac{n!}{(r-2)!(n-r+2)!}:\frac{n!}{(r-1)!(n-r+1)!}:\frac{n!}{r!(n-r)!}=1:3:5$

From here, we get ,

$\frac{r-1}{n-r+2}=\frac{1}{3}and\frac{r}{n-r+1}=\frac{3}{5}$

Which can be written as

$n-4r+5=0and3n-8r+3=0$

From these equations we get,

$n=7andr=3$

Question:11 Prove that the coefficient of $x^n$ in the expansion of $(1+x{)}^{2n}$ is twice the coefficient of $x^n$ in the expansion of $(1+x{)}^{2n-1}$.

Answer:

As we know that the general $(r+1{)}^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b{)}^n$ is given by

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

So, general $(r+1{)}^{th}$ term $T_{r+1}$ in the binomial expansion of $(1+x{)}^{2n}$ is,

$T_{r+1}{=}^{2n}{C}_r{1}^{2n-r}{x}^r$

$x^n$ will come when $r=n$,

So, Coefficient of $x^n$ in the binomial expansion of $(1+x{)}^{2n}$ is,

$K_{1x^n}{=}^{2n}{C}_n$

Now,

the general $(r+1{)}^{th}$ term $T_{r+1}$ in the binomial expansion of $(1+x{)}^{2n-1}$ is,

$T_{r+1}{=}^{2n-1}{C}_r{1}^{2n-1-r}{x}^r$

Here also $x^n$ will come when $r=n$,

So, Coefficient of $x^n$ in the binomial expansion of $(1+x{)}^{2n-1}$ is,

$K_{2x^n}{=}^{2n-1}{C}_n$

Now, As we can see

${}^{2n-1}{C}_n=\frac{(2n-1)!}{n!(2n-1-n)!}=\frac{(2n-1)!}{n!(n-1)!}=\frac{(2n)!}{2n(n!)(n-1)!}=\frac{(2n)!}{2(n!)(n!)}$

${}^{2n-1}{C}_n=\frac{1}{2}{\times }^{2n}{C}_n$

$2{\times }^{2n-1}{C}_n{=}^{2n}{C}_n$

$2\times K_{2x^n}=K_{1x^n}$

Hence, the coefficient of $x^n$ in the expansion of $(1+x{)}^{2n}$ is twice the coefficient of $x^n$ in the expansion of $(1+x{)}^{2n-1}$.

Question:12 Find a positive value of m for which the coefficient of $x^2$ in the expansion $(1+x{)}^m$ is 6.

Answer:

As we know that the general $(r+1{)}^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b{)}^n$ is given by

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

So, the general $(r+1{)}^{th}$ term $T_{r+1}$ in the binomial expansion of $(1+x{)}^m$ is

$T_{r+1}{=}^m{C}_r{1}^{m-r}{x}^r{=}^m{C}_r{x}^r$

$x^2$ will come when $r=2$. So,

The coeficient of $x^2$ in the binomial expansion of $(1+x{)}^m$ = 6

${\Rightarrow }^m{C}_2=6$

$\Rightarrow \frac{m!}{2!(m-2)!}=6$

$\Rightarrow \frac{m(m-1)}{2}=6$

$\Rightarrow m(m-1)=12$

$\Rightarrow m^2-m-12=0$

$\Rightarrow (m+3)(m-4)=0$

$\Rightarrow m=4or-3$

Hence the positive value of m for which the coefficient of $x^2$ in the expansion $(1+x{)}^m$ is 6, is 4.