Question

In the expansion of ${(3^{-\frac{x}{4}}+3^{\frac{5x}{4}})}^n$ the sum of binomial coefficient is $256$ and four times the term with greatest binomial coefficient exceeds the square of the third term by $21n$ then value of $x$ is

• A. $\frac{1}{2}$
• B. $1$
• C. $0$
• D. cannot be determined

Answer 1

Answer: (A)

$\frac{1}{2}$

Given,
The sum of all binomial coefficients is $256$
So, $2^n=256=2^8$ and $n=8$
The term with the greatest binomial coefficient is the middle term that is $T_5$
$T_5{=}_8^4\text{C}\left(3^{\frac{-x}{4}}{)}^4\right(3^{\frac{5x}{4}}{)}^4$
and third term, $T_3{=}_8^2\text{C}\left(3^{\frac{-x}{4}}{)}^6\right(3^{\frac{5x}{4}}{)}^2$
Given,
$4{(}_8^4\text{C}\left(3^{\frac{-x}{4}}{)}^4\right(3^{\frac{5x}{4}}{)}^4)-{(}_8^2\text{C}(3^{\frac{-x}{4}}{)}^6(3^{\frac{5x}{4}}{)}^2{)}^2=21.8=168$
$\Rightarrow {280.3}^{4x}-\frac{8.7}{2}{.3}^x=168$
$\Rightarrow {280.3}^{2x}{3}^{2x}-\frac{8.7}{2}{.3}^x=168$
Let, $3^{2x}=y$
So,
$10y^2-28y-6=0$
$\Rightarrow y=3$ or $-\frac{1}{5}$
It cannot be negative, so $y=3$
$3^{2x}=3$

Equating powers, we get
$2x=1\Rightarrow x=\frac{1}{2}$