JEE Main 2023 (Online) 11th April Evening ShiftQuestion 1
The sum of the coefficients of three consecutive terms in the binomial expansion of $(1+x)^{n+2}$, which are in the ratio $1:3:5$, is equal to
Binomial Theorem · Properties and Sums of Binomial Coefficients
Solution
Concept. Two ratios of consecutive coefficients give two linear equations in $N$ and $k$. Formulas used. (1) $\dfrac{{}^{N}C_{k+1}}{{}^{N}C_k}=\dfrac{N-k}{k+1}$Step 1 - Set up. Let the coefficients be ${}^{N}C_k,{}^{N}C_{k+1},{}^{N}C_{k+2}$ with $N=n+2$. Step 2 - First ratio $3$. By (1), $\dfrac{N-k}{k+1}=3$, so $N=4k+3$. Step 3 - Second ratio $\frac53$. $\dfrac{N-k-1}{k+2}=\dfrac53$, so $3N=8k+13$. Step 4 - Solve. $3(4k+3)=8k+13$ gives $k=1$, $N=7$. The coefficients are $7,21,35$ (ratio $1:3:5$). Step 5 - Sum. $7+21+35=63$. Answer: (C)