Question

If the coefficients of x⁷ and x⁸ in ${\left(2+\frac{x}{3}\right)}^n$ are equal, then n is
(a) 56
(b) 55
(c) 45
(d) 15

Answer 1

$$\begin{gathered} \mathrm{In}{(a+x)}^n \\ {T}_{r+1}{=}^n{C}_r{a}^{n-r}{x}^r \\ \therefore \mathrm{In}{\left(2+\frac{x}{3}\right)}^n \end{gathered}$$
Coefficient of x⁷ is
$$\begin{gathered} T_8={}^n{C}_7{\left(2\right)}^{n-7}{\left(\frac{x}{3}\right)}^7 \\ {T}_8={}^n{C}_7\frac{2^{n-7}{x}^7}{3^7} \end{gathered}$$
and coefficient of x⁸ is
$$\begin{gathered} T_9={}^n{C}_8{\left(2\right)}^{n-8}{\left(\frac{x}{3}\right)}^8 \\ {T}_9={}^n{C}_8\frac{2^{n-8}{x}^8}{3^8} \end{gathered}$$
Since T₈ = T₉ i.e coefficient of T₈ = coefficient of T₉
$$\begin{gathered} {\Rightarrow }^n{C}_7\frac{2^{n-7}}{3^7}=\frac{{}^n{C}_8{2}^{n-8}}{3^8} \\ \Rightarrow \frac{n!2^{n-7}}{7!(n-7)!\times 3^7}=\frac{n!\times 2^{n-8}}{8!(n-8)!\times 3^8} \\ \mathrm{i}.\mathrm{e}.\frac{8!}{7!}\frac{(n-8)!}{(n-7)(n-8)!}=\frac{2^{n-8}}{3\times 2^{n-7}} \\ \mathrm{i}.\mathrm{e}.\frac{8}{n-7}=\frac{1}{6}\Rightarrow 48=n-7 \\ \mathrm{i}.\mathrm{e}.n=55 \end{gathered}$$

Hence, the correct answer is option B.