JEE Main 2025 — Mathematics Algebra

$\displaystyle \begin{aligned} & \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k^3+6 k^2+11 k+5}{(k+3)!} \\ & =\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k^3+6 k^2+11 k+6-1}{(k+3)!} \\ & =\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{(k+1)(k+2)(k+3)-1}{(k+3)!} \\ & =\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{(k+1)(k+2)(k+3)}{(k+3)!}-\frac{1}{(k+3)!} \\ & =\lim _{k=1} \sum_{k=1}^n\left(\frac{1}{k!}-\frac{1}{(k+3)!}\right)\end{aligned}$​n→∞lim​k=1∑n​(k+3)!k3+6k2+11k+5​=n→∞lim​k=1∑n​(k+3)!k3+6k2+11k+6−1​=n→∞lim​k=1∑n​(k+3)!(k+1)(k+2)(k+3)−1​=n→∞lim​k=1∑n​(k+3)!(k+1)(k+2)(k+3)​−(k+3)!1​=k=1lim​k=1∑n​(k!1​−(k+3)!1​)​ $\displaystyle \begin{aligned} & =\lim _{k=1}\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!} \ldots+\frac{1}{n!}-\frac{1}{4!}-\frac{1}{5!}-\frac{1}{6!} \ldots .-\frac{1}{(n+3)!}\right) \\ & \quad=\frac{1}{1}+\frac{1}{2}+\frac{1}{6}=\frac{10}{6}=\frac{5}{3}\end{aligned}$​=k=1lim​(1!1​+2!1​+3!1​+4!1​…+n!1​−4!1​−5!1​−6!1​….−(n+3)!1​)=11​+21​+61​=610​=35​​