JEE PYQ: Binomial Theorem Question 8
Question 8 - 2021 (18 Mar Shift 1)
Let $(1 + x + 2x^2)^{20} = a_0 + a_1 x + a_2 x^2 + \ldots + a_{40} x^{40}$. Then $a_1 + a_3 + a_5 + \ldots + a_{37}$ is equal to:
(1) $2^{20}(2^{20} - 21)$ (2) $2^{19}(2^{20} - 21)$ (3) $2^{19}(2^{20} + 21)$ (4) $2^{20}(2^{20} + 21)$
Show Answer
Answer: (2)
Solution
Using $x = 1$ and $x = -1$, and subtracting $a_{39}$: $a_1 + a_3 + \ldots + a_{37} = 2^{19}(2^{20} - 21)$.
BETA
