Mathematics · 2026
JEE Main · 8 April 2026, Shift 2 · Q8
If 26(2^3/3(^12 C_2)+2^5/5(^12 C_4)+2^7/7(^12 C_6)+⋯+(2^13)/13(^12 C_12))=3^13-α, then α is equal to:
If $\displaystyle 26\left(\frac{2^3}{3}\left({ }^{12} \mathrm{C}_2\right)+\frac{2^5}{5}\left({ }^{12} \mathrm{C}_4\right)+\frac{2^7}{7}\left({ }^{12} \mathrm{C}_6\right)+\cdots+\frac{2^{13}}{13}\left({ }^{12} \mathrm{C}_{12}\right)\right)=3^{13}-\alpha$, then $\displaystyle \alpha$ is equal to:
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle 51$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.