Class 11 Maths Exercise 8.2 Solutions

It is known that \( ({r}+1)^{{n}} \) term, \( \left({T}_{{r}+1}\right) \), in the binomial expression of \( ({a}+ b) { }^{n} \) is given by
\({T}_{{r}+1}={ }^{{n}} {C}_{{r}} {a}^{{n}-{r}} {b}^{{r}}\)
Assuming that \(x^{5}\) occurs in the \( ({r}+1)^{{n}} \) term of the expression \( ({x}+3)^{8} \), we obtain
\({T}_{{r}+1}={ }^{8} {C}_{{r}}({x})^{8-{r}}(3)^{{r}}\)
Comparing the indices of \(x\) in \( {x}^{5} \) in \( {T}_{{r}+1} \)
We, obtain \( r=3 \)
Thus, the coefficient of \( x^5 \) is \( { }^{8} {C}_{3}(3)^{3}=\frac{8!}{3!5!} \times 3^{3}=\frac{8.7.6 . 5!}{3.2 .5!} . 3^{3}=1512 \)

It is known that \( ({r}+1)^{{n}} \) term \( \left({T}_{{r}+1}\right) \), in the binomial expression 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 \( (r+1)^{12} \) term of the expression \( (a- 2 b)^{12} \), we obtain
\({T}_{{r}+1}={ }^{12} {C}_{{r}}({a})^{12-{r}}(-2 {b})^{{r}}={ }^{12} {C}_{{r}}({a})^{12-{r}}({b})^{{r}}\)
Comparing the indices of \(a\) and \(b\) in \( {a}^{5} {b}^{7} \) in \( {T}_{{r}+1} \)
We, obtain \( r=7 \)
Thus, the coefficient of \( {a}^{5} {~b}^{7} \) is
\({ }^{12} {C}_{{r}}(-2)^{7}=\frac{12!}{7!5!} \cdot 2^{7}=\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7!}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 7!} \cdot(-2)^{7}=-(792)(128)=-101376\)

It is known that the general term \( {T}_{{r}+1} \) \(\{\)which is the \( ({r}+1)^{{n}} \) term\(\}\) in the binomial expression 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
\(T^{r+1}={ }^{6} C_{r}\left(x^{2}\right)^{6-r}(-y)^{r}=(-1)^{r} \ { }^{6} C_{r} \cdot x^{12-2 r} \cdot y^{r}\)

It is known that the general term \( {T}_{{r}+1} \) \(\{\)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 \( \left(x^{2}-y x\right)^{12} \) is
\({T}_{{r}+1}={ }^{12} {C}_{{r}}\left({x}^{2}\right)^{12-{r}}(-{y} {x})^{{r}}\)
\(=(-1)^{{r} } \ { }^{12} {C}_{{r}} \cdot {x}^{24-2 {r}} \cdot {y}^{{r}}\)
\(=(-1)^{{r}}{ }^{12} {C}_{{r}} \cdot {x}^{24-{r}} \cdot {y}^{{r}}\)

It is known \( ({r}+1)^{{n}} \) term, \( {T}_{{r}+1} \) in the binomial expansion of \( ({a}+{b}) ^{{n}}\) is given by \( {T}_{{r}+1}={ }^{n} {C}_{{r}} {a}^{{n}-{r}} {b}^{{r}} \)
Thus, the \( 4^{\text {th}} \) term is the expansion of \( \left(x^{2}-2 y\right)^{12} \) is
\({T}_{4}={T}_{3+1}={ }^{12} {C}_{3}({x})^{12-3}(-2 {y})^{3}\)
\(=(-1)^{3} \cdot \frac{12!}{3!9!} \cdot {x}^{9} \cdot(2)^{3} \cdot {y}^{3}\)
\(=\frac{12 \cdot 11 \cdot 10}{3 \cdot 2} \cdot(2)^{3} {x}^{9} {y}^{3}\)
\(=-1760 {x}^{9} {y}^{3}\)

It is known \( ({r}+1)^{{n}} \) term, \( {T}_{{r}+1} \) in the binomial expansion of \( ({a}+{b})^{{n}} \) is given by \( {T}_{{r}+1}={ }^{{n}} {C}_{{r}} {a}^{{n}-{r}} b^{{r}} \)
Thus, the \( 13^{\text {th}} \) term in the expansion of \( \left\{9 x-\frac{1}{3 \sqrt{x}}\right\}^{18} \) is
\({T}_{13}={T}_{12-1}={ }^{18} {C}_{12}(9 {x})^{18-12}\left\{-\frac{1}{3 \sqrt{x}}\right\}^{12}\)
\(=(-1)^{12} \frac{18!}{12!6!}(9)^{6}(x)^{6}\left(\frac{1}{3}\right)^{12} x\left(\frac{1}{\sqrt{x}}\right)^{12}\)
\(=\frac{18.17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12!}{12!6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} \cdot {x}^{6} \frac{1}{x^{6}} \cdot 3^{12} \frac{1}{3^{12}}\)
\(=18564\)

It is known that in the expansion of \( ({a}+{b})^{{n}} \) in \( n \) is odd, then there are two middle terms
Namely \( \left(\frac{n+1}{2}\right)^{n} \) term and \( \left(\frac{n+1}{2}+1\right)^{n} \) term.
Therefore, the middle terms in the expansion \( \left(3-\frac{x^{3}}{6}\right)^{7} \) are \( \left(\frac{7+1}{2}\right)^{n}=4^{\text {th}} \) and \( \left(\frac{7+1}{2}+1\right)^{n}=5^{\text {th}} \) term.
\({T}_{4}={T}_{3+1}={ }^{7} {C}_{3}(3)^{7-3}-\frac{x^{3}}{6}=(-1)^{3} \frac{7!}{3!4!} \cdot 3^{4} \cdot \frac{x^{9}}{6^{3}}\)
\(=\frac{7 \cdot 6 \cdot 5 \cdot 4!}{3 \cdot 2 \cdot 4!} \cdot 3^{4} \cdot \frac{1}{2^{3} \cdot 3^{3}} \cdot {x}^{9}=-\frac{105}{8} {x}^{4}\)
\({~T}_{5}={T}_{4+1}={ }^{7} {C}_{4}(3)^{7-4}\left(-\frac{x^{3}}{6}\right)^{4}=(-1)^{4} \frac{7!}{4!3!} \cdot 3^{3} \cdot \frac{x^{12}}{6^{4}}\)
\(=\frac{7 \cdot 6 \cdot 5 \cdot 4!}{4!3 \cdot 2} \cdot \frac{3^{3}}{2^{4} \cdot 3^{4}} \cdot x^{12}=\frac{35}{48} x^{12}\)
Thus, the middle terms in the expansion of \( \left(3-\frac{x^{3}}{6}\right)^{7} \) are \( -\frac{105}{8} {x}^{9} \) and \( \frac{35}{48} {x}^{12} \)

It is known that in the expansion of \( ({a}+{b})^{n} \), in \(n \) is even the middle term is \( \left(\frac{n}{2}+1\right)^{n} \) term.
Therefore, the middle term in the expansion of \( \left\{\frac{x}{3}+9 y\right\}^{10} \) is \( \left(\frac{10}{2}+1\right)^{n}= \) \( 6^{\text {th }} \)
\({T}_{4}={T}_{5+1}={ }^{10} {C}_{5}\left(\frac{x}{3}\right)^{10-5}(9 {y})^{5}=\frac{10!}{5!5!} \cdot \frac{x^{5}}{3^{6}} \cdot 9^{5} \cdot {y}^{5}\)
\(=\frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5!}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 5!} \cdot \frac{1}{3^{6}} \cdot 3^{10} \cdot x^{5} y^{5}\left[9^{5}=\left(3^{2}\right)^{5}=3^{10}\right]\)
\(=252 \times 3^{6} \cdot {x}^{5} \cdot {y}^{5}=6123 {x}^{5} {y}^{5} \)
Thus, the middle term in the expansion of \( \left(\frac{x}{3}+9 y\right)^{10} \) is \( 6123 x^{5} y^{5} \)

It is known that \( ({r}+1)^{{n}} \) 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 \( a^{n}\) occurs in the \( (r+1)^{{n}} \) term of the expansion \( (1+a)^{m+n} \), we obtain
\({T}_{{r}+1}={}^{{m}+{n}}{C}_{{r}}(1)^{{m}+{n}-{r}}({a})^{{r}}={ }^{{m}+{n}} {C}_{{r}} {a}^{{r}}\)
Comparing the indices of a in \( {a}^{{n}} \) in \( {T}_{{r}+1} \)
We, obtain \( {r}={m} \)
Therefore, the coefficient of an is
\( { }^{{M}+n} {C}_{{r}}=\frac{(m+n)!}{m!(m+n-m)!}=\frac{(m!+n!)}{m!n!} \ldots (1)\)
Assuming that an occurs in the \( ({k}+1)^{{n}} \) 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 \( {T}_{{k}+1} \)
We, obtain
\( {K}={n} \)
Therefore, the coefficient of an is
\( { }^{{M}+n} {C}_{{n}}=\frac{(m+n)!}{n!(m+n-n)!}=\frac{(m+n)!}{n!m!} \ldots (2)\)
Thus, from (1) and (2), it can be observed that the coefficient of \( {a}^{{n}} \) in the expansion of \( (1+a)^{m+n} \) is equal.

It is known that \( ({k}+1)^{{n}} \) term \( \left({T}_{{k}+1}\right) \) in the binomial expansion of \( ({a}+ b)^{n}\) is given by \( {T}_{{k}+1}={ }^{{n}} {C}_{{k}} {a}^{{n}-{k}} {b}_{{k}} \)
Therefore, \( ({r}-1)^{{n}} \) term in the expansion of \( ({x}+1)^{1} \) is
\( {T}_{{r}-1}={ }^{{n}} {C}_{{r}-2}({x})^{{n}-({r}-2)}(1)^{({r}-2)}={ }^{{n}} {C}_{{r}-2} {x}^{{n}-{r}-2} \)
\( ({r}+1) \) term in the expansion of \( ({x}+1)^{{n}} \) is
\( {T}_{{r}-1}={ }^{{n}} {C}_{{r}}({x})^{{n}-{r}}(1)^{{r}}={ }^{{n}} {C}_{1} {x}^{{n}-{r}} \)
\( {r}^{\text {th}} \) term in the expansion of \( ({x}+1)^{{n}} \) is
Therefore, the coefficient of the \( ({r}-1)^{\text {th}} , {r}^{\text {th}} \) and \( ({r}+1)^{\text {th}} \) term in the expansion of \( ({x}+1)^{{n}} \)
\( { }^{{n}} {c}_{{r-2}},{ }^{{n}} {c}_{{r-1}} \), and \({}^{n}c_{r}\) are respectively. Since these - coefficient are in the ratio \( 1: 3: 5 \), we obtain
\(=\frac{n_{c_{r-2}}}{n_{c_{r-1}}}=\frac{1}{3} \text { and } \frac{n_{c_{r-1}}}{n_{c_{r}}}=\frac{3}{5}\)
\(\frac{n_{c_{r-2}}}{n_{c_{r-1}}}=\frac{n!}{(r-1)!(n-r+1)} \times \frac{r!(n-r)!}{n!}=\frac{(r-1)!(r-2)!9 n-r+1)}{(r-2)!(n-r+1)!(n-r+2)!}\)
\(=\frac{r}{n-r+2}\)
\(\therefore \frac{r-1}{n-r+2}=\frac{1}{3}\)
\(=3 {r}-3={n}-{r}+2\)
\(={n}-4 {r}+5=0 \ldots(1)\)
\(\frac{n_{c_{r-1}}}{n_{c_{r}}}=\frac{n!}{(r-1)!(n-r+1)} \times \frac{r!(n-r)!}{n!}=\frac{r(r-1)(n-r)!}{(r-1)!(n-r+1)(n-r)!}\)
\(=\frac{r}{n-r+1}\)
\(\therefore \frac{r}{n-r+1}=\frac{3}{5}\)
\(=5 r=3 n-3 r+3\)
\(=3 n-8 r+3=0 \ldots(2)\)
Multiplying (1) by 3 and subtracting it from (2), we obtain
\(4 r-12=0\)
\(=r=3\)
Putting the value of \( r \) in (1), we obtain \( n \)
\( -12+5=0 \)
\(-{n}=7\)
thus, \( {n}=7 \) and \( {r}=3 \)

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 \({x}^{{n}}\) occurs in the \( ({r}+1)^{\text {th}} \) term of the expansion of \( (1+{x}) { }^{2 n} \), we obtain
\({T}_{{r}+1}={ }^{2 {n}} {c}_{{r}}(1)^{2 {n}-{r}}({x})^{{r}}={}^{2n}c_{{r}}({x})^{{r}}\)
Comparing the indices of \(x\) in \( {x}^{{n}} \) and in \( {T}_{{r}+2} \), we obtain \( {r}={n} \)
Therefore, the coefficient of \( x^{n} \) in the expansion of \( (1+x)^{2 n} \) is
\({ }^{2 {n}} {c}_{{n}}=\frac{(2 n)!}{n!(2 n-n)!}=\frac{(2 n)!}{n!n!}=\frac{(2 n)!}{n!^{2}} \ldots (1)\)
Assuming that \({x}^{{n}}\) occurs in the \( ({k}+1)^{\text {th}} \) term of the expansion of \( (1+ x) { }^{2 n-2} \), we obtain
\({T}_{{k}+1}={ }^{2 {n}} {c}_{{k}}(1)^{2 {n}-{r}-{k}}({x})^{{k}=2 {n}} {c}_{{k}}({x})^{{k}}\)
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 \( (1+x)^{2 n-1} \) is
\({}^{2 {n}-1} c_{n}=\frac{(2 n-1)!}{n!(2 n-1-n)!}=\frac{(2 n-1)!}{n!(n-1)!}\)
\(=\frac{2 n(2 n-1)!}{2 n \cdot n!(n-1)!}=\frac{(2 n)!}{2 n!n!}=\frac{1}{2}\left[\frac{(2 n)!}{(n!)^{2}}\right] \ldots (2)\)
From (1) and (2), it is observed that
\(\frac{1}{2}\left({ }^{2 n} c_{r}\right)={ }^{2 n-1} c_{n}\)
\(={ }^{2 n} c_{n}=2\left({ }^{2 n-1} c_{n}\right)\)
Therefore, the coefficient of \( x^{n} \) expansion of \( (1+x)^{2 n} \) is twice the coefficient of \( x^{n} \) in the expansion of \( (1+x)^{2 n-1} \)
Hence, proved.

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 \( {x}^{2} \) occurs in the \( ({r}+1)^{\text {th}} \) term of the expansion of \( (1+{x})^{n}\), we obtain
\({T}_{{r}+1}={ }^{{n}} {C}_{{r}}(1)^{{n}-{r}}({x})^{{r}}={ }^{{n}} {c}_{{r}}({x})^{{r}}\)
Comparing, the coefficient of \(x\) in \( {x}^{2} \) and in \( {T}_{{r}+1} \), we obtain \( {r}=2 \)
Therefore, the coefficient of \( x^{2} \) is \( { }^{n} c_{2} \)
It is given that the coefficient of \(x^{2}\) in the expansion \( (1+{x})^{n} \) is \(6 \).
\(={ }^{{n}} {C}_{2}=6\)
\(=\frac{m!}{2!(m-2)!}=6\)
\(=\frac{m(m+1)(m-2)!}{2 \times(m-2)!}=6\)
\(={m}({m}-1)=12\)
\(={m}^{2}-{m}-12=0\)
\(={m}^{2}-4 {m}+3 {m}-12=0\)
\(={m}({m}-4)+3({m}-4)=0\)
\(=({m}-4)({m}+3)=0\)
\( =({m}-4)=0 \) or \( ({m}+3)=0 \)
\( ={m}=4 \) 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 \(4 \).