Question

In the expansion of $(2+\frac{x}{3}{)}^n$, coefficients of $x^7$ and $x^8$ are equal. Then $n=$

• A. $49$
• B. $50$
• C. $55$
• D. $56$

Answer 1

The correct option is C $55$

We know, $T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$
Applying to the above question, we get
$T_{r+1}={}^n{C}_r{2}^{n-r}{3}^{-r}{x}^r$
Applying the given conditions, i.e. coefficient of $x^7$ is equal to coefficient of $x^8$
Therefore ${}^n{C}_7{2}^{n-7}{3}^{-7}={}^n{C}_8{2}^{n-8}{3}^{-8}$
$\Rightarrow 6{}^n{C}_7={}^n{C}_8$
$\Rightarrow \frac{6.\left(n\right)!}{(n-7)!7!}=\frac{\left(n\right)!}{(n-8)!8!}$
$\Rightarrow 48=n-7$
$\Rightarrow n=55$