Question

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

• A. 56
• B. 55
• C. 49
• D. 50

Answer 1

Answer: (B)

Since $T_{r+1}{=}^n{C}_r{a}^{n-r}{x}^r$ in expansion of ${\left(a+x\right)}^n$
$T_8={}^n{C}_7{\left(2\right)}^{n-7}{\left(\frac{x}{3}\right)}^7={}^n{C}_7\frac{2^{n-7}}{3^7}{x}^7$
and $T_9={}^n{C}_8{\left(2\right)}^{n-8}{\left(\frac{x}{3}\right)}^8={}^n{C}_8\frac{2^{n-8}}{3^8}{x}^8$
Therefore, ${}^n{C}_7\frac{2^{n-7}}{3^7}={}^n{C}_8\frac{2^{n-8}}{3^8}$
(since it is given that coefficient of $x^7$ = coefficient $x^8$)
$\Rightarrow \frac{n!}{7!\left(n-7\right)!}\times \frac{8!\left(n-8\right)!}{n!}=\frac{2^{n-8}}{3^8}.\frac{3^7}{2^{n-7}}$
$\Rightarrow \frac{8}{n-7}=\frac{1}{6}\Rightarrow n=55$