NCERT Solutions for Class 11 Maths Chapter 8: Binomial Theorem - Exercise 8.1

Exercise 8.2

1. Find the coefficient of ${x^5}$ in ${(x + 3)^n}$

Ans: It is known that ${(r + 1)^th}$ term, $\left( {{T_{r + 1}}} \right).$ in the binomial expansion of ${(a + b)^n}$ is given by ${{\text{T}}_{r + 1}}{ = ^ n }{{\text{C}}_{\text{r}}}{{\text{a}}^{{\text{n}} - {\text{r}}}}{{\mathbf{b}}^r}$

Assuming that ${x^2}$ occurs in the ${(r + 1)^n}$ term of the expansion ${(x + 3)^8}$, we obtain ${T_{r + 1}}{ = ^a}{C_r}{(x)^{8 - r}}{(3)^r}$

Comparing the indices of $x$ in ${x^5}$ in ${T_{r + 1}}$. We obtain $r = 3$ Thus, the coefficient of ${x^5}$ is $^8{C_3}{(3)^3} = \dfrac{{8!}}{{3!5!}} \times {3^3} = \dfrac{{8 \cdot 7 \cdot 6 \cdot 5!}}{{3 \cdot 2.5!}} * {3^3} = 1512$.

2. Find the coefficient of ${{\text{a}}^5}\;{{\text{b}}^{\text{7}}}$ in ${({\text{a}} - 2\;{\text{b}})^{12}}$

Ans: It is known that ${(r + 1)^th}$ term, $\left( {{T_{r + 2}}} \right)$, in the binomial expansion of ${(a + b)^n}$ is given by ${T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r}$

Assuming that ${a^5}{b^7}$ occurs in the ${(r + 1)^{1 = }}$ term of the expansion ${(a - 2b)^{22}}$, we obtain

${T_{r + 1}}{ = ^{12}}{C_r}{(a)^{2 - r}}{( - 2b)^r}{ = ^{12}}{C_r}{( - 2)^r}{(a)^{12 - r}}{(b)^r}$

Comparing the indices of a and $b$ in ${a^5}{b^7}$ in ${T_{r + 2}}$. We obtain $r = 7$ Thus, the coefficient of ${a^5}{b^7}$ is $^2{{\text{C}}_1}{( - 2)^7} = \dfrac{{12!}}{{7!5!}} \cdot {2^7} = \dfrac{{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7!}}{{5 \cdot 4 \cdot 3 \cdot 2 - 7!}} \cdot {( - 2)^7} =  - (792)(128) =  - 101376$

3. Write the general term in the expansion of ${\left( {{x^2} - y} \right)^6}$

Ans: It is known that the general term ${T_{r + 1}}\left\{ {} \right.$ which is the ${(r + 1)^n}$ term $\} $ in the binomial expansion of ${(a + b)^n}$ is given by ${T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r}.$

Thus, the general term in the expansion of $\left( {{x^2} - {y^6}} \right)$ is ${T_{r + 1}}{ = ^6}{C_r}{\left( {{x^2}} \right)^{6 - r}}{( - y)^r} = {( - 1)^{r}}{{ ^6}C_r} - {x^{12 - 2r}} \cdot {y^r}$

4. Write the general term in the expansion of ${\left( {{x^2} - yx} \right)^{12}},x \ne 0$

Ans: It is known that the general term ${T_{r + 1}}\left\{ {} \right.$ which is the ${(r + 1)^th}$ term $\} $ in the binomial expansion of ${(a + b)^n}$ is given by ${T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r}.$

Thus, the general term in the expansion of ${\left( {{x^2} - yx} \right)^{12}}$ is ${T_{r + 1}}{ = ^{12}}{C_r}{\left( {{x^2}} \right)^{12 - t}}{( - yx)^r} = {( - 1)^{{r}}}{^12}{C_r} - {x^{24 - 2r}} - {y^r} = {( - 1)^{r - 2}}{C_r}{x^{24 - r}} \cdot {y^r}$

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

Ans: It is known ${(r + 1)^n}$ term, ${T_{r + 1}}$. in the binomial expansion of ${(a + b)^n}$ is given by ${{\text{T}}_{{\text{r}} + 1}}{ = ^n}{{\text{C}}_r}{{\mathbf{a}}^{n - r}}\;{{\text{b}}^r}.$

Thus, the ${4^{{\text{th }}}}$ term in the expansion of ${\left( {{x^2} - 2y} \right)^{12}}$ is ${T_4} = {T_{3 + 1}}{ = ^{12}}{C_3}{(x)^{12 - 3}}{( - 2y)^3} = {( - 1)^3} \cdot \dfrac{{12!}}{{319!}} \cdot {x^9} \cdot {(2)^3} - {y^3} =  - \dfrac{{12 \cdot 11 \cdot 10}}{{3 \cdot 2}} \cdot {(2)^3}{x^9}{y^3} =  - 1760{x^9}{y^3}$

6. Find the ${13^{{\text{th }}}}$ term in the expansion of ${\left( {9x - \dfrac{1}{{3\sqrt x }}} \right)^{18}},x \ne 0$

Ans: It is known ${({\text{r}} + 1)^{\text{th}}}$ term, ${T_{r + 1}}$, in the binomial expansion of ${({\text{a}} + {\text{b}})^n}$ is given by ${{\text{T}}_{\varepsilon  + 1}}{ = ^n}{{\text{C}}_r}{{\text{a}}^{n - 1}}\;{{\text{b}}^r}$

Thus, the ${13^{{\text{th }}}}$ term in the expansion of ${\left( {9x - \dfrac{1}{{3\sqrt x }}} \right)^{18}}$ is ${T_{13}} = {T_{12 + 1}}{ = ^{18}}{C_{12}}{(9x)^{18 - 12}}{\left( { - \dfrac{1}{{3\sqrt x }}} \right)^{12}}$

$ = {( - 1)^{12}}\dfrac{{18!}}{{12!6!}}{(9)^6}{(x)^6}{\left( {\dfrac{1}{3}} \right)^{12}}{\left( {\dfrac{1}{{\sqrt x }}} \right)^{12}}$

$ = \dfrac{{18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13.12!}}{{12! \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}} \cdot {x^6}\left( {\dfrac{1}{{{x^6}}}} \right) \cdot {3^{12}}\left( {\dfrac{1}{{{3^{12}}}}} \right)\quad \left[ {{9^6} = {{\left( {{3^2}} \right)}^6} = {3^{12}}} \right]$

$ = 18564$

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

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

Therefore, the middle terms in the expansion ${\left( {3 - \dfrac{{{x^3}}}{6}} \right)^7}\operatorname{are} {\left( {\dfrac{{7 + 1}}{2}} \right)^{\text{th}}} = {4^{{\text{th }}}}$ and ${\left( {\dfrac{{7 + 1}}{2} + 1} \right)^{{\text{th }}}} = {5^{{\text{th }}}}$ term ${T_4} = {T_{3 + 1}}{ = ^7}{C_2}{(3)^{7 - 3}}{\left( { - \dfrac{{{x^3}}}{6}} \right)^3} = {( - 1)^3}\dfrac{{7!}}{{3!4!}} - {3^4} \cdot \dfrac{{{x^2}}}{{{6^3}}}$

$ =  - \dfrac{{7 \cdot 6 \cdot 5.4!}}{{3 \cdot 2.4!}} \cdot {3^4} \cdot \dfrac{1}{{{2^3} \cdot {3^3}}} \cdot {x^9} =  - \dfrac{{105}}{8}{x^9 }$

${{\text{T}}_{\text{5}}} = {{\text{T}}_{4 + 1}}{ = ^7}{{\text{C}}_4}{(3)^{7 - 4}}{\left( { - \dfrac{{{{\text{x}}^3}}}{6}} \right)^4} = {( - 1)^4}\dfrac{{71}}{{4!3!}} \cdot {3^3} \cdot \dfrac{{{{\text{x}}^{22}}}}{{{6^4}}}$

$ = \dfrac{{7 \cdot 6 \cdot 5.4!}}{{4! \cdot 3 \cdot 2}} \cdot \dfrac{{{3^3}}}{{{2^4} \cdot {3^4}}} \cdot {x^{12}} = \dfrac{{35}}{{48}}{x^{12}}$

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

8.  Find the middle terms in the expansion of ${\left( {\dfrac{x}{3} + 9y} \right)^{10}}$

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

Therefore, the middle term in the expansion of ${\left( {\dfrac{x}{3} + 9y} \right)^{10}}$ is ${\left( {\dfrac{{10}}{2} + 1} \right)^n} = {6^{{\text{th }}}}$ ${T_4} = {T_{5 + 1}}{ = ^{30}}{C_5}{\left( {\dfrac{x}{3}} \right)^{10 - 5}}{(9y)^5} = \dfrac{{10!}}{{5!5!}} \cdot \dfrac{{{x^5}}}{{{3^5}}} \cdot {9^5} \cdot {y^5}$

$ = \dfrac{{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6.5!}}{{5 \cdot 4 \cdot 3 \cdot 2.5!}} \cdot \dfrac{1}{{{3^5}}} \cdot {3^{10}} \cdot {x^2}{y^3}$

$\left[ {{9^5} = {{\left( {{3^2}} \right)}^5} = {3^{10}}} \right]$

$ = 252 \times {3^5} \cdot {x^5} \cdot {y^5} = 6123{x^5}{y^5}$

Thus, the middle term in the expansion of ${\left( {\dfrac{x}{3} + 9y} \right)^{10}}$ is $61236{x^5}{y^5}$.

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

Ans: It is known that ${(r + 1)^{th}}$ term, $\left( {{T_{r + 1}}} \right)$, in the binomial expansion of ${(a + b)^n}$ is given by ${T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r}$

Assuming that a" occurs in the ${(r + 1)^{{\text{th }}}}$ term of the expansion ${(1 + a)^{n + m}}$, we obtain ${T_{r + 1}}{ = ^{m + n}}{C_r}{(1)^{n + m  - r}}{(a)^r}{ = ^{m + n}}{C_r}{a^r}$

Comparing the indices of a in ${a^n}$ in ${T_{r + 1}}$. We obtain $r = m$ Therefore, the coefficient of ${a^n}$ is $^{m + n}{C_m} = \dfrac{{(m + n)!}}{{m!(m + n - m)!}} = \dfrac{{(m + n)!}}{{m!n!}}$

Assuming that ${a^n}$ occurs in the ${(k + 1)^{{\text{th }}}}$ term of the expansion ${(1 + a)^{m + n}}$, we obtain ${T_{k + 1}}{ = ^{m + n}}{C_k}{(1)^{m + n - k}}{(a)^k}{ = ^{m + n}}{C_k}{(a)^k}$

Comparing the indices of a in ${a^n}$ and in ${T_{k - 1}}$. We obtain ${\mathbf{k}} = {\mathbf{n}}$

Therefore, the coefficient of ${{\mathbf{a}}^{\text{n}}}$ is $^{m + n}{{\text{C}}_n} = \dfrac{{(m + n)!}}{{n!(m + n - n)!}} = \dfrac{{(m + n)!}}{{n!m!}} \ldots  \ldots .(2)$

Thus, from (1) and (2), it can be observed that the coefficients of ${a^m}$ and ${a^n}$ in the expansion of ${(1 + a)^{m + n}}$ are equal.

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 $r$.

Ans: It is known that ${(k + 1)^{th}}$ term, $\left( {{T_{k + 1}}} \right)$, in the binomial expansion of ${(a + b)^n}$ is given by ${T_{k + 1}}{ = ^n}{{\mathbf{C}}_k}{{\mathbf{a}}^{n - k}}{{\mathbf{b}}^k}$

Therefore, ${(r - 1)^{{\text{n }}}}$ term in the expansion of ${(x + 1)^n }$ is 

$(r + 1)$ term in the expansion of ${(x + 1)^n}$ is ${T_{r - 1}}{ = ^n}{C_r}{(x)^{n - r}}{(1)^r}{ = ^n}{C_r}{x^{n - r}}$

${{\mathbf{r}}^{\text{e}}}$ term in the expansion of ${({\text{x}} + 1)^{\text{n}}}$ is ${T_r}{ = ^n}{C_{r - 1}}{(x)^{n - 1r - 1}}{(1)^{(r - 2)}}{ = ^n}{C_{r - 1}}{x^{n - r + 1}}$

Therefore, the coefficients of the ${(r - 1)^{{\text{th }}}},{r^{{\text{th }}}}$ and ${(r + 1)^{{\text{th }}}}$ terms in the expansion of ${(x + 1)^n}$ $^n{{\text{C}}_{r - 2}}{,^n}{{\text{C}}_{r - 1}}$, and $^n{{\text{C}}_r}$ are respectively. Since these coefficients are in the ratio 1: 3: 5, we obtain $\dfrac{{^n{C_{r - 2}}}}{{^n{C_{r - 1}}}} = \dfrac{1}{3}$ and $\dfrac{{^n{C_{r - 1}}}}{{^n{C_r}}} = \dfrac{3}{5}$

$\dfrac{{^n{C_{r - 2}}}}{{^n{C_{r - 1}}}} = \dfrac{{n!}}{{(r - 2)!(n - r + 2)!}} \times \dfrac{{(r - 1)!(n - r + 1)!}}{{n!}} = \dfrac{{(r - 1)(r - 2)!(n - r + 1)!}}{{(r - 2)!(n - r + 2)!(n - r + 1)!}}$

$ = \dfrac{{{\mathbf{r}} - 1}}{{{\mathbf{n}} - {\mathbf{r}} + 2}}$

$\therefore \dfrac{{{\mathbf{r}} - 1}}{{{\mathbf{n}} - {\mathbf{r}} + 2}} - \dfrac{1}{3}$

$ \Rightarrow 3x - 3 = n - r + 2$

$ \Rightarrow n - 4r + 5 = 0 \ldots .(1)$

$\dfrac{{^n{{\text{C}}_{r - 1}}}}{{^n{{\text{C}}_r}}} = \dfrac{{{\text{n}}!}}{{({\text{r}} - 1)!({\text{n}} - {\text{r}} + 1)}} \times \dfrac{{{{\text{r}}!}({\text{n}} - {\text{r}})!}}{{{\text{n}}!}} = \dfrac{{{\text{r}}({\text{r}} - 1)!({\text{n}} - {\text{r}})!}}{{({\text{r}} - 1)!({\text{n}} - {\text{r}} + 1)({\text{n}} - {\text{r}})!}}$

$ = \dfrac{{\mathbf{r}}}{{{\mathbf{n}} - {\mathbf{r}} + {\mathbf{1}}}}$

$\therefore \dfrac{r}{{n - r + 1}} = \dfrac{3}{5}$

$ \Rightarrow 5r = 3n - 3r + 3$

$ \Rightarrow 3n - 8r + 3 = 0$

Multiplying (1) by 3 and subtracting it from (2), we obtain $4r - 12 = 0$

$ \Rightarrow {\mathbf{r}} = 3$

Putting the value of $r$ in $(1)$, we obtain $n$ $ - 12 + 5 = 0$

$ \Rightarrow {\text{n}} = 7$

Thus, $n = 7$ and ${\mathbf{r}} = 3$

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 ${(1 + x)^{2n - 1}}$.

Ans: It is known that ${(r + 1)^{{\text{th }}}}$ term, $\left( {{T_{r + 1}}} \right)$, in the binomial expansion of ${(a + b)^n}$ is given by ${T_{r + 1}}{ = ^n}{{\text{C}}_r}{{\text{a}}^{n - r}}\;{{\text{b}}^r}$

Assuming that ${x^n}$ occurs in the ${(r + 1)^n}$ term of the expansion of ${(1 + x)^{2n}}$, we obtain ${T_{r + 1}}{ = ^{2n}}{{\text{C}}_r}{(1)^{2n - r}}{({\text{x}})^r}{ = ^{2n}}{{\text{C}}_{\text{r}}}{({\text{x}})^r}$

Comparing the indices of $x$ in ${x^n}$ and in ${T_{r + 1}}$, we obtain $r = n$ Therefore, the coefficient of ${x^n}$ in the expansion of ${(1 + x)^{2n}}$ is $^{2n}{{\text{C}}_n} = \dfrac{{(2n)!}}{{n!(2n - n)!}} = \dfrac{{(2n)!}}{{n!n!}} = \dfrac{{(2n)!}}{{{{(n!)}^2}}} \ldots  \ldots .(1)$

Assuming that ${x^n}$ occurs in the ${(k + 1)^{{\text{in }}}}$ term of the expansion of ${(1 + x)^{2n - 2}}$, we obtain

Comparing the indices of $x$ in ${x^n}$ and in ${T_{k + 1}}$, we obtain $k = n$ Therefore, the coefficient of ${x^n}$ in the expansion of ${(1 + x)^{2n - 1}}$ is $^{2n - 1}{{\text{C}}_n} = \dfrac{{(2n - 1)!}}{{n!(2n - 1 - n)!}} = \dfrac{{(2n - 1)!}}{{n!(n - 1)!}}$

$ = \dfrac{{2n \cdot (2n - 1)!}}{{2n \cdot n!(n - 1)!}} = \dfrac{{(2n)!}}{{2n!n!}} = \dfrac{1}{2}\left[ {\dfrac{{(2n)!}}{{{{(n!)}^2}}}} \right] \ldots  \ldots .(2)$

From (1) and (2), it is observed that $\dfrac{1}{2}\left( {^{2n}{C_n}} \right){ = ^{2n - 1}}{C_n}$

${ \Rightarrow ^{2n}}{{\text{C}}_n} = 2\left( {^{2n - 1}{{\text{C}}_n}} \right)$

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

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

Ans: It is known that ${(r + 1)^{{\text{th }}}}$ term, $\left( {{T_{r + 1}}} \right)$, in the binomial expansion of ${(a + b)^n}$ is given by ${T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r}$

Assuming that ${{\text{x}}^2}$ occurs in the ${({\text{r}} + 1)^{\text{n}}}$ term of the expansion of ${(1 + {\text{x}})^{\text{n}}}$, we obtain ${T_{r + 1}}{ = ^n}{C_r}{(1)^{n - r}}{(x)^r}{ = ^m}{C_n}{(x)^r}$

Comparing the indices of $x$ in ${x^2}$ and in ${T_{r + 1}}$, we obtain $r = 2$ Therefore, the coefficient of ${x^2}$ is $^m{{\text{C}}_2}$ It is given that the coefficient of ${x^2}$ in the expansion ${(1 + x)^m}$ is 6 . ${\therefore ^m}{{\text{C}}_2} = 6$

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

$ \Rightarrow \dfrac{{m(m - 1)(m - 2)!}}{{2 \times (m - 2)!}} = 6$

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

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

$ \Rightarrow {m^2} - 4m + 3m - 12 = 0$

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

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

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

$ \Rightarrow m = 4$ or $m =  - 3$

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

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

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