Question

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

• A. $45$
• B. $55$
• C. $56$
• D. $13$

Answer 1

The correct option is C $56$

Since $T_{r+1}{=}^n{C}_r{a}^{n-r}{x}^r$ in expansion of ${(a+x)}^n$

$\therefore T_8{=}^n{C}_7{3}^{n-7}{\left(\frac{x}{2}\right)}^7$

${=}^n{C}_7{3}^{n-7}{x}^7{2}^{-7}$

Coefficient of $x^7{=}^n{C}_7{3}^{n-7}{2}^{-7}$

$\therefore T_9{=}^n{C}_8{3}^{n-8}{\left(\frac{x}{2}\right)}^8$

${=}^n{C}_8{3}^{n-8}{x}^8{2}^{-8}$

Coefficient of $x^8{=}^n{C}_8{3}^{n-8}{2}^{-8}$

Given:Coefficient of $x^7=$Coefficient of $x^8$

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

$\Rightarrow {}^n{C}_{7}3=\frac{{}^n{C}_8}{2}$

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

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

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

$\Rightarrow 48=(n-7)$

$\Rightarrow 48+7=n$

$\Rightarrow n=55$