Question

If the co-efficient of $x^7$ and $x^8$ in ${(2+\frac{x}{3})}^n$ are equal, then n is-

• A. $56$
• B. $55$
• C. $45$
• D. $15$

Answer 1

Answer: (B)

$55$

Given,

$(2+\frac{x}{3}{)}^n$

We know that, $T_{r+1}{=}^n{C}_r.2^{n-r}.(\frac{x}{3}{)}^r$. . . . .(1)

If, $r=7$,

$T_8{=}^n{C}_7.\frac{2^{n-7}}{3^7}$

If, $r=8$

$T_9{=}^n{C}_8.\frac{2^{n-8}}{3^8}$

From the question,

$T_8=T_9$

${}^n{C}_7.\frac{2^{n-7}}{3^7}{=}^n{C}_8.\frac{2^{n-8}}{3^8}$

$\frac{n!}{7!(n-7)!}.\frac{2^{n-7}}{3^7}=\frac{n!}{8!(n-8)!}.\frac{2^{n-8}}{3^8}$

$\frac{1}{n-7}.2=\frac{1}{3\times 8}$

$\frac{2}{n-7}=\frac{1}{24}$

$n-7=48$

$n=7+48=55$

The correct option is B.