Mathematics · 2026
JEE Main · 22 January 2026, Shift 1 · Q6
The coefficient of x^48 in (1+x)+2(1+x)^2+3(1+x)^3+…+100(1+x)^100 is equal to
The coefficient of $\displaystyle x^{48}$ in $\displaystyle (1+x)+2(1+x)^2+3(1+x)^3+\ldots+100(1+x)^{100}$ is equal to
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle 100 \cdot{ }^{101} \mathrm{C}_{49}-{ }^{101} \mathrm{C}_{50}$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.