Question

Let $T_r$ denote the $r^{th}$ term in the expansion of ${\left(2^x+\frac{1}{4^x}\right)}^n.$ If the ratio $T_3:T_2=7:1$ and the sum of the coefficients of the second and third term is $36,$ then the value of $x$ is

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $-\frac{1}{4}$
• D. $-\frac{1}{3}$

Answer 1

Answer: (D)

It is given that $\frac{T_3}{T_2}=7\Rightarrow \frac{{}^n{C}_2{\left(2^x\right)}^{n-2}{\left(\frac{1}{4^x}\right)}^2}{{}^n{C}_1{\left(2^x\right)}^{n-1}{\left(\frac{1}{4^x}\right)}^1}=7$
$\Rightarrow \left(\frac{n-1}{2}\right)2^{-3x}=7$
Also, ${}^n{C}_1+{}^n{C}_2=36$
$\Rightarrow n+\frac{n\left(n-1\right)}{2}=36$
$\Rightarrow n^2+n-72=0\Rightarrow n=8,-9$
$n=8\Rightarrow \frac{7}{2}{2}^{-3x}=7$
$\Rightarrow 2^{-3x-1}=1=2^0$
$\Rightarrow -3x-1=0\Rightarrow x=-\frac{1}{3}$