Mathematics · 2024
JEE Main · 8 April 2024, Shift 1 · Q24
Let α=Σ_r=0^n(4 r^2+2 r+1)^n C_r and β=(Σ_r=0^n (^n C_r)/(r+1))+1/(n+1). If 140<(2 α)/(β)<281, then the value of n is ____.
Let $\displaystyle \alpha=\sum_{r=0}^n\left(4 r^2+2 r+1\right){ }^n C_r$ and $\displaystyle \beta=\left(\sum_{r=0}^n \frac{{ }^n C_r}{r+1}\right)+\frac{1}{n+1}$. If $\displaystyle 140<\frac{2 \alpha}{\beta}<281$, then the value of $\displaystyle n$ is $\displaystyle \_\_\_\_$.
Official answer
From NTA’s final answer key for this paper.
5
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.