Question

The coefficient of $x^7$ and $x^8$ in the expansion of ${(2+\frac{x}{3})}^n$ are equal, then $n$ is equal to-

• A. $35$
• B. $45$
• C. $55$
• D. $Noneofthese$

Answer 1

The correct option is C $55$

${[2+\frac{x}{3}]}^n$

${\left(2\right)}^n{[1+\frac{x}{6}]}^n$

${[1+\frac{x}{6}]}^n$

coefficient of $x^7={{}^{n}C}_7{{\left(\frac{x}{6}\right)}^7}^{}={{}^{n}C}_7{\left(\frac{1}{6}\right)}^7{\left(x\right)}^7$

coefficient of $x^8={{}^{n}C}_8{\left(\frac{x}{6}\right)}^8={{}^{n}C}_8{\left(\frac{1}{6}\right)}^8{\left(x\right)}^8$

given

${\left(2\right)}^n{{}^{n}C}_7{\left(\frac{1}{6}\right)}^7={\left(2\right)}^n{{}^{n}C}_8{\left(\frac{1}{6}\right)}^8$

${{}^{n}C}_7=\frac{1}{6}{{}^{n}C}_8$

$\frac{n!}{(n-7)!7!}=\frac{1}{6}\frac{n!}{(n-8)!8!}$

$8\times 6(n-7)!7!=(n-7)(n-8)!7!$

$48=n-7\Rightarrow n=48+7=55\Rightarrow n=55$