JEE Main 2020 (Online) 2nd September Evening ShiftQuestion 1
For a positive integer $n$, $\left(1+\frac1x\right)^n$ is expanded in increasing powers of $x$. If three consecutive coefficients in this expansion are in the ratio, $2:5:12$, then $n$ is equal to ________.
Binomial Theorem · Properties and Sums of Binomial Coefficients
Answer$118$
Solution
Concept. In increasing powers of $x$, the coefficients of $\left(1+\frac1x\right)^n$ come in the order ${}^{n}C_n,{}^{n}C_{n-1},\dots,{}^{n}C_0$. Formulas used. (1) $\dfrac{{}^{n}C_k}{{}^{n}C_{k+1}}=\dfrac{k+1}{n-k}$Step 1 - Set up. Three consecutive coefficients are ${}^{n}C_{k+2},{}^{n}C_{k+1},{}^{n}C_k$, in the ratio $2:5:12$. Step 2 - Two ratios. By (1), $$\frac{{}^{n}C_{k+1}}{{}^{n}C_{k+2}}=\frac{k+2}{n-k-1}=\frac52\ \Longrightarrow\ 7k=5n-9,$$ $$\frac{{}^{n}C_k}{{}^{n}C_{k+1}}=\frac{k+1}{n-k}=\frac{12}{5}\ \Longrightarrow\ 17k=12n-5.$$ Step 3 - Solve. $17(5n-9)=7(12n-5)$ gives $85n-153=84n-35$, so $n=118$ and $k=83$. (Check: $\frac{85}{34}=\frac52$ and $\frac{84}{35}=\frac{12}{5}$.) Answer: 118