JEE Main 2021 (Online) 25th July Evening ShiftQuestion 1
Let $n\in\mathbf{N}$ and $[x]$ denote the greatest integer less than or equal to $x$. If the sum of $(n+1)$ terms ${}^{n}C_0,\ 3\cdot{}^{n}C_1,\ 5\cdot{}^{n}C_2,\ 7\cdot{}^{n}C_3,\ \ldots$ is equal to $2^{100}\cdot101$, then $2\left[\frac{n-1}{2}\right]$ is equal to $\_\_\_\_\_\_$ .
Binomial Theorem · Properties and Sums of Binomial Coefficients
Answer$98$
Solution
Concept. Split $(2k+1)\,{}^{n}C_k$ into $2k\,{}^{n}C_k$ and ${}^{n}C_k$. Formulas used. (1) $\sum k\,{}^{n}C_k=n\,2^{n-1}$ and $\sum{}^{n}C_k=2^n$Step 1 - The sum. The $(k+1)$-th term is $(2k+1)\,{}^{n}C_k$. By (1), $$\sum_{k=0}^{n}(2k+1)\,{}^{n}C_k=2n\,2^{n-1}+2^n=(n+1)\,2^n.$$ Step 2 - Find $n$. $(n+1)\,2^n=101\cdot2^{100}$ gives $n=100$. Step 3 - Value. $2\left[\frac{99}{2}\right]=2\cdot49=98$. Answer: 98