Mathematics · 2021
JEE Main · 4 August 2021, Shift 1 · Q65
If (1+x+x^2)^4=a_0+a_1 x+a_2 x^2+… …+a_8 x^8, then the value of a_0+4 a_2+16 a_4+….+2^8 a_8 is equal to
If $\displaystyle \left(1+x+x^2\right)^4=a_0+a_1 x+a_2 x^2+\ldots \ldots+a_8 x^8$, then the value of $\displaystyle a_0+4 a_2+16 a_4+\ldots .+2^8 a_8$ is equal to
Official answer
(3)
$\displaystyle 1241$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.