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 coefficients of $2nd$ and $3rd$ 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
$2^{3x}=\frac{1}{2}=2^{-1}$
$\Rightarrow 3x=-1\Rightarrow x=-\frac{1}{3}$