Mathematics · 2025
JEE Main · 3 April 2025, Shift 2 · Q21
Let (1+x+x^2)^10=a_0+a_1 x+a_2 x^2+…+a_20 x^20. If (a_1+a_3+a_5+…+a_19)-11 a_2=121 k, then k is equal to ____.
Let $\displaystyle \left(1+x+x^2\right)^{10}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$. If $\displaystyle \left(a_1+a_3+a_5+\ldots+a_{19}\right)-11 a_2=121 k$, then $\displaystyle k$ is equal to $\displaystyle \_\_\_\_$.
Official answer
From NTA’s final answer key for this paper.
239
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.