Mathematics · 2021
JEE Main · 18 March 2021, Shift 1 · Q66
Let (1+x+2 x^2)^20= a_0+ a_1 x+ a_2 x^2+….+ a_40 x^40. Then, a_1+ a_3+ a_5+…..+ a_37 is equal to:
Let $\displaystyle \left(1+x+2 x^2\right)^{20}=\mathrm{a}_0+\mathrm{a}_1 x+\mathrm{a}_2 x^2+\ldots .+\mathrm{a}_{40} x^{40}$. Then, $\displaystyle \mathrm{a}_1+\mathrm{a}_3+\mathrm{a}_5+\ldots . .+\mathrm{a}_{37}$ is equal to :
Official answer
(3)
$\displaystyle 2^{19}\left(2^{20}-21\right)$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.