JEE Main 2021 (Online) 24th February Morning ShiftQuestion 1
The value of $-{}^{15}C_{1}+2\cdot{}^{15}C_{2}-3\cdot{}^{15}C_{3}+\ldots-15\cdot{}^{15}C_{15}+{}^{14}C_{1}+{}^{14}C_{3}+{}^{14}C_{5}+\ldots+{}^{14}C_{11}$ is :
Binomial Theorem · Properties and Sums of Binomial Coefficients
Solution
Concept. The first part is $\sum(-1)^r r\,{}^{15}C_r$, which is $0$; the second is an odd-index sum with its last term missing. Formulas used. (1) $r\,{}^{n}C_r=n\,{}^{n-1}C_{r-1}$ and $\sum_{j}(-1)^j\,{}^{m}C_j=0$ $(m\ge1)$(2) ${}^{n}C_1+{}^{n}C_3+\cdots=2^{n-1}$Step 1 - First part. By (1), $$\sum_{r=1}^{15}(-1)^r\,r\,{}^{15}C_r=-15\sum_{j=0}^{14}(-1)^j\,{}^{14}C_j=0.$$ Step 2 - Second part. By (2), ${}^{14}C_1+{}^{14}C_3+\cdots+{}^{14}C_{13}=2^{13}$. Our sum stops at ${}^{14}C_{11}$, so remove ${}^{14}C_{13}=14$: $$2^{13}-14.$$ Step 3 - Total. $0+2^{13}-14=2^{13}-14$. Answer: (B)