Mathematics · 2025
JEE Main · 29 January 2025, Shift 1 · Q5
The value of lim_n → ∞(Σ_k=1^n (k^3+6 k^2+11 k+5)/((k+3)!)) is:
The value of $\displaystyle \lim _{n \rightarrow \infty}\left(\sum_{k=1}^n \frac{k^3+6 k^2+11 k+5}{(k+3)!}\right)$ is :
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle 5$/$\displaystyle 3$
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