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← JEE Main 2021 · 24 Feb · Shift 1
MathematicsSingle correct+4 / -1Question 64 of 90

JEE Main 2021 · 24 Feb · Shift 1, Mathematics question 64

The value of −15C1+2⋅15C2−3⋅15C3+⋯−15⋅15C15+14C1+14C3+14C5+⋯+14C11-{}^{15}C_1+2\cdot{}^{15}C_2-3\cdot{}^{15}C_3+\dots-15\cdot{}^{15}C_{15}+{}^{14}C_1+{}^{14}C_3+{}^{14}C_5+\dots+{}^{14}C_{11} is :

Choose one option
Answer and worked solution

Official answer: B

First part. ∑r=115(−1)rr(15r)=−15∑(−1)r−1(14r−1)=−15(1−1)14=0\sum_{r=1}^{15}(-1)^r r\binom{15}r=-15\sum(-1)^{r-1}\binom{14}{r-1}=-15(1-1)^{14}=0.

Second part. Odd binomials of 14 sum to 2132^{13}; dropping 14C13=14{}^{14}C_{13}=14 leaves 213−142^{13}-14.

(B)