Question

If in the expansion of $\left(2^{x}+\frac{1}{4^{x}}\right)^{n}, \frac{T_{3}}{T_{2}}=7$ and the sum of the binomial 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)

Since, $\frac{T_{3}}{T_{2}}=7$ (given)
$\Rightarrow \frac{{ }^{n} C_{2}\left(2^{x}\right)^{n-2}\left(4^{-x}\right)^{2}}{{ }^{n} C_{1}\left(2^{x}\right)^{n-1} \cdot\left(4^{-x}\right)}=7$
$\Rightarrow \left(\frac{n-1}{2}\right) \cdot \frac{1}{\left(2^{x}\right)^{3}}=7$
Also, ${ }^{n} C_{2}+{ }^{n} C_{1}=36$
$\Rightarrow \frac{n(n-1)}{2}+n=36$
$\Rightarrow n^{2}+n-72=0$
$\Rightarrow n=8,-9$
$n=-9$ is not possible as in Eq. (1),
$n-1$ should be positive.
Substituting $n=8$ in Eq. (1), we get
$ 2^{3 x}=\frac{1}{2}=2^{-1} $
$\Rightarrow 3 x=-1 \Rightarrow x=-\frac{1}{3}$