Question

If in the expansion of $(2^x+4^{-x}{)}^n,\frac{T_3}{T_2}=7$ and the sum of the coefficients of $2^{nd}$ and $3^{rd}$ terms is $36$,

then the value of $x$ is :

• A. $-\frac{1}{3}$
• B. $-\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

$-\frac{1}{3}$

$2^x=t$

${(t+t^{-2})}^n$

$\Rightarrow T_2={{}^{n}C}_1{t}^{n-1}.t^{-2}$

$={{}^{n}C}_1{t}^{n-3}$

$\Rightarrow T_3={{}^{n}C}_2{t}^{n-2}.t^{-4}$

$={{}^{n}C}_2{t}^{n-6}$

$\Rightarrow \frac{T_3}{T_2}=7$

$\Rightarrow \frac{\frac{n(n-1)}{2}.t^{n-6}}{n.t^{n-3}}$ $=\frac{(n-1)}{2}.t^{-3}=7$

As, $T_2+T_3=36$

$\Rightarrow n{t}^{n-3}+\frac{n(n-1)}{2}.t^{n-6}=36$

$\Rightarrow n{t}^{n-3}[1+\frac{(n-1)}{2}.t^{-3}]=36$

$\Rightarrow n{t}^{n-3}[1+7]=36$

$\Rightarrow n{t}^{n-3}=\frac{9}{2}$

$\Rightarrow n.2^{x(n-3)}=\frac{9}{2}$

$\Rightarrow n=9$ ; $x=\frac{-1}{3}$

Hence the answer is $\frac{-1}{3}.$