JEE Main 2023 — Binomial Theorem Question with Solution

The problem asks for the sum of the coefficients of three consecutive terms in the binomial expansion of $(1+x{)}^{n+2}$, which are in the ratio 1 : 3 : 5.

Given that the ratios of the coefficients are 1:3:5, we let the terms be $T_r$, $T_{r+1}$, and $T_{r+2}$. The coefficients of these terms are ${}^{n+2}{C}_{r-1}$, ${}^{n+2}{C}_r$, and ${}^{n+2}{C}_{r+1}$, respectively.

$$\begin{array}{rl}& \frac{T_{r+1}}{T_r}=\frac{{}^{n+2}{C}_r}{{}^{n+2}{C}_{r-1}}=\frac{n+2-r+1}{r}=\frac{n+3-r}{r}=3\\ & \\ & n-4r+3=\mathrm{0......}(1)\\ & \\ & \frac{T_{r+2}}{T_{r+1}}=\frac{{}^{n+2}{C}_{r+1}}{{}^{n+2}{C}_r}=\frac{(n+2)-(r+1)+1}{r+1}=\frac{n-r+2}{r+1}=\frac{5}{3}\\ & \\ & 3n-8r+1=\mathrm{0......}(2)\end{array}$$

By solving (1) and (2), we get

$\Rightarrow n=5,r=2$

$$\begin{array}{rl}{T}_r+T_{r+1}+T_{r+2}& ={}^7{C}_1+{}^7{C}_2+{}^7{C}_3\\ & \\ & =7+21+35=63\end{array}$$