11th NCERT/CBSE Binomial Theorem Exercise 8.2 Questions 14
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Question (1)

Find the coefficient of x5 in (x + 3)8

Solution

It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by ${T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r}$
Assuming that x5 occurs in the (r + 1)th term of the expansion (x + 3)8, we obtain ${T_{r + 1}}{ = ^8}{C_r}{\left( x \right)^{8 - r}}{\left( 3 \right)^r}$
Comparing the indices of x in x5 and in Tr +1, we obtain r = 3
Thus, the coefficient of x5 is \[^8{C_3}{\left( 3 \right)^3} = \frac{{8!}}{{3!5!}} \times {3^3} = \frac{{8 \cdot 7 \cdot 6 \cdot 5!}}{{3 \cdot 2 \cdot 5!}} \cdot {3^3} = 1512\]

Question (2)

Find the coefficient of a5b7 in (a – 2b)12

Solution

It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by ${T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r}$
Assuming that a5b7 occurs in the (r + 1)th term of the expansion (a – 2b)12, we obtain
\[{T_{r + 1}}{ = ^{12}}{C_r}{\left( a \right)^{12 - r}}{\left( { - 2b} \right)^r}{ = ^{12}}{C_r}{\left( { - 2} \right)^r}{\left( a \right)^{12 - r}}{\left( b \right)^r}\] Comparing the indices of a and b in a5b7 and in Tr+1, we obtain r = 7
Thus, the coefficient of a5b7 is \[^{12}{C_7}{\left( { - 2} \right)^7} = - \frac{{12!}}{{7!5!}} \cdot {2^7} = - \frac{{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7!}}{{5 \cdot 4 \cdot 3 \cdot 2 \cdot 7!}} \cdot {2^7}\] \[ = - \left( {792} \right)\left( {128} \right) = - 101376\]

Question (3)

Write the general term in the expansion of (x2 – y)6

Solution

It is known that the general term Tr+1 {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 (x2 – y)6 \[{T_{r + 1}}{ = ^n}{C_r}{\left( {{x^2}} \right)^{6 - r}}{\left( { - y} \right)^r} = {\left( { - 1} \right)^6}{C_r}{x^{12 - 2r}}{y^r}\]

Question (4)

Write the general term in the expansion of (x2 – yx)12, x ≠ 0

Solution

It is known that the general term Tr+1 {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 (x2 – yx)12 is \[{T_{r + 1}}{ = ^{12}}{C_r}{\left( {{x^2}} \right)^{12 - r}}{\left( { - yx} \right)^r}\] \[ = {\left( { - 1} \right)^r}{\;^{12}}{C_r}{x^{24 - 2r}}{y^r}{x^r}\] \[ = {\left( { - 1} \right)^r}{\;^{12}}{C_r}{x^{24 - r}}{y^r}\]

Question (5)

Find the 4th term in the expansion of (x – 2y)12 .

Solution

It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by ${T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r}$ Thus, the 4th term in the expansion of (x – 2y)12 is \[{T_4} = {T_{3 + 1}}{ = ^{12}}{C_3}{\left( x \right)^{12 - 3}}{\left( { - 2y} \right)^3}\] \[ = {\left( { - 1} \right)^3} \cdot \frac{{12!}}{{3!9!}} \cdot {x^9} \cdot {\left( 2 \right)^3} \cdot {y^3}\] \[ = - \frac{{12 \cdot 11 \cdot 10}}{{3 \cdot 2}} \cdot {\left( 2 \right)^3}{x^9}{y^3}\] \[ = - 1760{x^9}{y^3}\]

Question (6)

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

Solution

It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by ${T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r}$
Thus, 13th term in the expansion of ${\left( {9x - \frac{1}{{3\sqrt x }}} \right)^{18}}$ is
\[{T_{13}} = {T_{12 + 1}}{ = ^{18}}{C_{12}}{\left( {9x} \right)^{18 - 12}}{\left( { - \frac{1}{{3\sqrt x }}} \right)^{12}}\] \[ = {\left( { - 1} \right)^{12}}\frac{{18!}}{{12!6!}}{\left( 9 \right)^6}{\left( x \right)^6}{\left( {\frac{1}{3}} \right)^{12}}{\left( {\frac{1}{{\sqrt x }}} \right)^{12}}\] \[ = \frac{{18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12!}}{{12!6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}}{x^6}\left( {\frac{1}{{{x^3}}}} \right){3^{12}}\left( {\frac{1}{{{3^{12}}}}} \right)\] Note $\left[ {{9^6} = {{\left( {{3^2}} \right)}^6} = {3^{12}}} \right]$ \[ = 18564\]

Question (7)

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, ${\left( {\frac{{n + 1}}{2}} \right)^{th}}$ term and ${\left( {\frac{{n + 1}}{2} + 1} \right)^{th}}$ term.
Therefore, the middle terms in the expansion of ${\left( {3 - \frac{{{x^3}}}{6}} \right)^7}$ are ${\left( {\frac{{7 + 1}}{2}} \right)^{th}} = {4^{th}}$ term and ${\left( {\frac{{7 + 1}}{2} + 1} \right)^{th}} = {5^{th}}$ term. \[{T_4} = {T_{3 + 1}}{ = ^7}{C_4}{\left( 3 \right)^{7 - 3}}{\left( { - \frac{{{x^3}}}{6}} \right)^3}\] \[ = {\left( { - 1} \right)^3}\frac{{7!}}{{3!4!}} \cdot {3^4} \cdot \frac{{{x^9}}}{{{6^3}}}\] \[ = {\left( { - 1} \right)^3}\frac{{7!}}{{3!4!}} \cdot {3^4} \cdot \frac{{{x^9}}}{{{6^3}}}\] \[ = - \frac{{7 \cdot 6 \cdot 5 \cdot 4!}}{{4! \cdot 3 \cdot 2}} \cdot {3^4} \cdot \frac{1}{{{2^3} \cdot {3^3}}} \cdot {x^9} = - \frac{{105}}{8}{x^9}\] \[{T_5} = {T_{4 + 1}}{ = ^7}{C_4}{\left( 3 \right)^{7 - 4}}{\left( { - \frac{{{x^3}}}{6}} \right)^4}\] \[ = {\left( { - 1} \right)^4}\frac{{7!}}{{4!3!}}{\left( 3 \right)^3} \cdot \frac{{{x^{12}}}}{{{6^4}}}\] \[ = \frac{{7 \cdot 6 \cdot 5 \cdot 4!}}{{4! \cdot 3 \cdot 2}} \cdot \frac{{{3^3}}}{{{2^4} \cdot {3^4}}} \cdot {x^{12}} = \frac{{35}}{{48}}{x^{12}}\] Thus, the middle terms in the expansion of ${\left( {3 - \frac{{{x^3}}}{6}} \right)^7}$ are $ - \frac{{105}}{8}{x^9}$ and $[\frac{{35}}{{48}}{x^{12}}$

Question (8)

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 ${\left( {\frac{n}{2} + 1} \right)^{th}}$ term. Therefore, the middle term in the expansion of ${\left( {\frac{x}{3} + 9y} \right)^{10}}$ is ${\left( {\frac{{10}}{2} + 1} \right)^{th}} = {6^{th}}$ \[{T_6} = {T_{5 + 1}}{ = ^{10}}{C_5}{\left( {\frac{x}{3}} \right)^{10 - 5}}\left( {9{y^5}} \right)\] \[ = \frac{{10!}}{{5!5!}} \cdot \frac{{{x^5}}}{{{3^5}}} \cdot {9^5} \cdot {y^5}\] \[ = \frac{{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5!}}{{5 \cdot 4 \cdot 3 \cdot 2 \cdot 5!}} \cdot \frac{1}{{{3^5}}} \cdot {9^5} \cdot {y^5}\] Note : $\left[ {{9^5} = {{\left( {{3^2}} \right)}^5} = {3^{10}}} \right]$ \[ = 252 \times {3^5} \cdot {x^5} \cdot {y^5} = 6123{x^5}{y^5}\] Thus, the middle term in the expansion of ${\left( {\frac{x}{3} + 9y} \right)^{10}}$ is 61236 x5y5.

Question (9)

In the expansion of (1 + a)m + n, prove that coefficients of am and an are equal.

Solution

It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by ${T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r}$
Assuming that am occurs in the (r + 1)th term of the expansion (1 + a)m + n, we obtain \[{T_{r + 1}}{ = ^{m + n}}{C_r}{\left( 1 \right)^{m + n - r}}{\left( a \right)^r}{ = ^{m + n}}{C_r}{a^r}\] Comparing the indices of a in am and in Tr + 1, we obtain r = m
Therefore, the coefficient of am is
\[^{m + n}{C_m} = \frac{{\left( {m + n} \right)!}}{{m!\left( {m + n - m} \right)!}} = \frac{{\left( {m + n} \right)!}}{{m!n!}} - - - (i)\] Assuming that an occurs in the (k + 1)th term of the expansion (1 + a)m+n, we obtain \[{T_{k + 1}}{ = ^{m + n}}{C_k}{\left( 1 \right)^{m + n - k}}{\left( a \right)^k}{ = ^{m + n}}{C_k}{\left( a \right)^k}\] Comparing the indices of a in an and in Tk + 1, we obtain k = n
Therefore, the coefficient of an is
\[^{m + n}{C_n} = \frac{{\left( {m + n} \right)!}}{{n!\left( {m + n - n} \right)!}} = \frac{{\left( {m + n} \right)!}}{{n!m!}} - - - (ii)\] Thus, from (1) and (2), it can be observed that the coefficients of am and an in the expansion of (1 + a)m + n are equal.

Question (10)

The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n are in the ratio 1:3:5. Find n and r.

Solution

It is known that (k + 1)th term, (Tk+1), in the binomial expansion of (a + b)n is given by ${T_{k + 1}}{ = ^n}{C_k}{a^{n - k}}{b^k}$
Therefore, (r – 1)th term in the expansion of (x + 1)n is
\[{T_{r - 1}}{ = ^n}{C_{r - 2}}{\left( x \right)^{n - \left( {r - 2} \right)}}{\left( 1 \right)^{\left( {r - 2} \right)}}{ = ^n}{C_{r - 2}}{x^{n - r + 2}}\] rth term in the expansion of (x + 1)n is
\[{T_r}{ = ^n}{C_{r - 1}}{\left( x \right)^{n - \left( {r - 1} \right)}}{\left( 1 \right)^{\left( {r - 1} \right)}}{ = ^n}{C_{r - 1}}{x^{n - r + 1}}\] (r + 1)th term in the expansion of (x + 1)n is \[{T_{r + 1}}{ = ^n}{C_r}{\left( x \right)^{n - r}}{\left( 1 \right)^r}{ = ^n}{C_r}{x^{n - r}}\] Therefore, the coefficients of the (r – 1)th, rth, and (r + 1)th terms in the expansion of (x + 1)n are respectively. Since these coefficients are in the ratio 1:3:5, we obtain $\frac{{^n{C_{r - 2}}}}{{^n{C_{r - 1}}}} = \frac{1}{3}$ and $\frac{{^n{C_{r - 1}}}}{{^n{C_r}}} = \frac{3}{5}$ \[\frac{{^n{C_{r - 2}}}}{{^n{C_{r - 1}}}} = \frac{{n!}}{{\left( {r - 2} \right)!\left( {n - r + 2} \right)!}} \times \frac{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}{{n!}}\] \[ = \frac{{\left( {r - 1} \right)\left( {r - 2} \right)!\left( {n - r + 1} \right)!}}{{\left( {r - 2} \right)!\left( {n - r + 2} \right)\left( {n - r + 1} \right)!}} = \frac{{r - 1}}{{n - r + 2}}\] \[\therefore \frac{{r - 1}}{{n - r + 2}} = \frac{1}{3}\] \[ \Rightarrow 3r - 3 = n - r + 2\] \[ \Rightarrow n - 4r + 5 = 0 - - - (1)\] \[\frac{{^n{C_{r - 1}}}}{{^n{C_r}}} = \frac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)}} \times \frac{{r!\left( {n - r} \right)!}}{{n!}}\] \[ = \frac{{\left( {r - 1} \right)\left( {r - 2} \right)!\left( {n - r + 1} \right)!}}{{\left( {r - 2} \right)!\left( {n - r + 2} \right)\left( {n - r + 1} \right)!}}\] \[ = \frac{r}{{n - r + 1}}\] \[ \therefore \frac{r}{{n - r + 1}} = \frac{3}{5}\] \[ \Rightarrow 5r = 3n - 3r + 3\] \[ \Rightarrow 3n - 8r + 3 = 0 - - - (2)\] Multiplying (1) by 3 and subtracting it from (2), we obtain
4r – 12 = 0
⇒ r = 3
Putting the value of r in (1), we obtain
n – 12 + 5 = 0 ⇒ n = 7 Thus, n = 7 and r = 3

Question (11)

Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1 .

Solution

It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by ${T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r}$
Assuming that xn occurs in the (r + 1)th term of the expansion of (1 + x)2n, we obtain
Comparing the indices of x in xn and in Tr + 1, we obtain r = n Therefore, the coefficient of xn in the expansion of (1 + x)2n is \[^{2n}{C_n} = \frac{{\left( {2n} \right)!}}{{n!\left( {2n - n} \right)!}} = \frac{{\left( {2n} \right)!}}{{n!n!}} = \frac{{\left( {2n} \right)!}}{{{{\left( {n!} \right)}^2}}} - - - (1)\] Assuming that xn occurs in the (k +1)th term of the expansion (1 + x)2n – 1, we obtain \[{T_{k + 1}}{ = ^{2n - 1}}{C_k}{\left( 1 \right)^{2n - 1 - k}}{\left( x \right)^k}{ = ^{2n - 1}}{C_k}{\left( x \right)^k}\] Comparing the indices of x in xn and Tk + 1, we obtain k = n
Therefore, the coefficient of xn in the expansion of (1 + x)2n –1 is \[^{2n - 1}{C_n} = \frac{{\left( {2n - 1} \right)!}}{{n!\left( {2n - 1 - n} \right)!}} = \frac{{\left( {2n - 1} \right)!}}{{n!\left( {n - 1} \right)!}}\] \[ = \frac{{2n\left( {2n - 1} \right)!}}{{2n \cdot n!\left( {n - 1} \right)!}} = \frac{{\left( {2n} \right)!}}{{2 \cdot n!n!}} = \frac{1}{2}\left[ {\frac{{\left( {2n} \right)!}}{{{{\left( {n!} \right)}^2}}}} \right] - - - (2)\] From (1) and (2), it is observed that \[\frac{1}{2}\left( {^{2n}{C_n}} \right){ = ^{2n - 1}}{C_n}\] \[{ \Rightarrow ^{2n}}{C_n} = 2\left( {^{2n - 1}{C_n}} \right)\] Therefore, the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1. Hence, proved.

Question (12)

Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6.

Solution

It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by ${T_{r + 1}}{ = ^m}{C_r}{a^{n - r}}{b^r}$
Assuming that x2 occurs in the (r + 1)th term of the expansion (1 +x)m, we obtain
\[{T_{r + 1}}{ = ^m}{C_r}{\left( 1 \right)^{m - r}}{\left( x \right)^r}{ = ^m}{C_r}{\left( x \right)^r}\] Comparing the indices of x in x2 and in Tr + 1, we obtain r = 2 Therefore, the coefficient of x2 is. It is given that the coefficient of x2 in the expansion (1 + x)m is 6.
\[ \therefore ^m{C_2} = 6\] \[ \Rightarrow \frac{{m!}}{{2!\left( {m - 2} \right)!}} = 6\] \[ \Rightarrow \frac{{m\left( {m - 1} \right)\left( {m - 2} \right)!}}{{2 \times \left( {m - 2} \right)!}} = 6\] \[ \Rightarrow m\left( {m - 1} \right) = 12\] \[ \Rightarrow {m^2} - m - 12 = 0\] \[ \Rightarrow {m^2} - 4m + 3m - 12 = 0\] \[ \Rightarrow m\left( {m - 4} \right) + 3\left( {m - 4} \right) = 0\] \[ \Rightarrow \left( {m - 4} \right)\left( {m + 3} \right) = 0\] \[ \Rightarrow \left( {m - 4} \right) = 0 \; \;or \; \;\left( {m + 3} \right) = 0\] \[ \Rightarrow m = 4 \; \; or \;\; m = - 3\] Thus, the positive value of m, for which the coefficient of x2 in the expansion (1 + x)m is 6, is 4.
Exercise 8.1 ⇐
⇒Exercise 8.3