Mathematics · 2026
JEE Main · 28 January 2026, Shift 2 · Q6
The sum of the coefficients of x^499 and x^500 in (1+x)^1000+x(1+x)^999+x^2(1+x)^998+…+x^1000 is:
The sum of the coefficients of $\displaystyle x^{499}$ and $\displaystyle x^{500}$ in $\displaystyle (1+x)^{1000}+x(1+x)^{999}+x^2(1+x)^{998}+\ldots+x^{1000}$ is :
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle { }^{1002} C_{500}$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.