JEE Main 2025 Jan 29 Shift 1, Mathematics Q15: Summation of Series

Factor the numerator: $k^3+6k^2+11k+6=(k+1)(k+2)(k+3)$, so

$k^3+6k^2+11k+5=(k+1)(k+2)(k+3)-1.$

Hence the summand splits as $\frac{(k+1)(k+2)(k+3)}{(k+3)!}-\frac{1}{(k+3)!}=\frac{1}{k!}-\frac{1}{(k+3)!}.$

The two pieces telescope when summed from $k=1$ to $\infty$:

$=\underset{n\to \infty }{\lim }\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots +\frac{1}{n!}-\frac{1}{4!}-\frac{1}{5!}-\frac{1}{6!}-\cdots -\frac{1}{(n+3)!}\right)$

= $\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}=1+\frac{1}{2}+\frac{1}{6}=\frac{10}{6}=\frac{5}{3}.$