Question

Let $T_{r}$ denote the $r^{t h}$ 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 $\frac{T_{3}}{T_{2}}=7\Rightarrow \frac{^{n} C_{2} \left(2^{x}\right)^{n - 2} \left(\frac{1}{4^{x}}\right)^{2}}{^{n} C_{1} \left(2^{x}\right)^{n - 1} \left(\frac{1}{4^{x}}\right)^{1}}=7$
$\Rightarrow \left(\frac{n - 1}{2}\right)2^{- 3 x}=7$
Also, $^{n}C_{1}+\,{}^{n}C_{2}=36$
$\Rightarrow \, n+\frac{n \left(n - 1\right)}{2}=36$
$\Rightarrow \, n^{2}+n-72=0\Rightarrow \, n=8, \, -9$
$n=8 \, \Rightarrow \, \frac{7}{2} \, 2^{- 3 x}=7$
$\Rightarrow 2^{- 3 x - 1}=1=2^{0}$
$\Rightarrow -3x-1=0\Rightarrow x=-\frac{1}{3}$