Question

In the expression of ${\left(2^x+\frac{1}{4^x}\right)}^n$ ratio of 2nd and third terms is given by$t_3/t_2=7$ and the sum of the co-efficients of 2nd and 3rd 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 $\frac{-1}{3}$

Consider the given expression.

$\Rightarrow {(2^x+\frac{1}{4^x})}^n$

Given:

$\frac{t_3}{t_2}=7$

By binomial expansion, we know that

$t_3={}^n{C}_2{\left(2^x\right)}^{n-2}{\left(\frac{1}{4^x}\right)}^2$

$t_2={}^n{C}_1{\left(2^x\right)}^{n-1}{\left(\frac{1}{4^x}\right)}^1$

Again, we have

${}^n{C}_1+{}^n{C}_2=36$

${}^{n+1}{C}_2=36$

$\frac{(n+1)!}{2!(n+1-2)!}=36$

$\frac{n(n+1)}{2}=36$

$n^2+n-72=0$

$n=8,-9$

Therefore,

$n=8$

Now,

$\frac{t_3}{t_2}=\frac{{}^n{C}_2{\left(2^x\right)}^{n-2}{\left(\frac{1}{4^x}\right)}^2}{{}^n{C}_1{\left(2^x\right)}^{n-1}{\left(\frac{1}{4^x}\right)}^1}$

$\frac{{}^8{C}_2{\left(2^x\right)}^{8-2}{\left(\frac{1}{4^x}\right)}^2}{{}^8{C}_1{\left(2^x\right)}^{8-1}{\left(\frac{1}{4^x}\right)}^1}=7$

$\frac{28{\left(2^x\right)}^6{\left(\frac{1}{{\left(2^2\right)}^x}\right)}^2}{8{\left(2^x\right)}^7{\left(\frac{1}{{\left(2^2\right)}^x}\right)}^1}=7$

$\frac{7\times 2^{6x-4x}}{2\times 2^{7x-2x}}=7$

$2^{2x-5x}=2$

$-3x=1$

$x=-\frac{1}{3}$

Hence, this is the required result.