Mathematics · 2023
JEE Main · 24 January 2023, Shift 2 · Q68
If (^30 C_1)^2+2(^30 C_2)^2+3(^30 C_3)^2+…+30(^30 C_30)^2=(α 60!)/((30!)^2) then α is equal to:
If $\displaystyle \left({ }^{30} \mathrm{C}_1\right)^2+2\left({ }^{30} \mathrm{C}_2\right)^2+3\left({ }^{30} \mathrm{C}_3\right)^2+\ldots+30\left({ }^{30} \mathrm{C}_{30}\right)^2=\frac{\alpha 60!}{(30!)^2}$ then $\displaystyle \alpha$ is equal to :
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 15$
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JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.