Question

If in the expansion of ${(2^x+\frac{1}{4^x})}^n,$ the ratio of the third term and the second term is $7$ and the sum of the coefficients of the second term and the third term 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

The correct option is A $-\frac{1}{3}$

$\frac{T_3}{T_2}=7$
$\Rightarrow \frac{{}^n{C}_2(2^x{)}^{n-2}\cdot {\left(4^{-x}\right)}^2}{{}^n{C}_1(2^x{)}^{n-1}\cdot \left(4^{-x}\right)}=7$
$\Rightarrow \left(\frac{n-1}{2}\right)\cdot \frac{1}{{\left(2^x\right)}^3}=7\cdots \left(1\right)$

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. $\left(1\right),$ $n-1$ should be positive.
Substituting $n=8$ in Eq. $\left(1\right),$ we get
$2^{3x}=\frac{1}{2}=2^{-1}$
$\Rightarrow 3x=-1$
$\Rightarrow x=-\frac{1}{3}$