Let S=Σn=1^∞(3n^2+8n+2)/((n+2)!). Find the value of S.

Write the numerator in terms of factorial-friendly pieces: $3n^2+8n+2=3(n+2)(n+1)-(n+2)-2$. Check: $3(n^2+3n+2)-n-2-2=3n^2+8n+2$. Dividing by $(n+2)!$, the term equals $\frac{3}{n!}-\frac{1}{(n+1)!}-\frac{2}{(n+2)!}$. Now $\sum_{n\ge1}\frac1{n!}=e-1$, $\sum_{n\ge1}\frac1{(n+1)!}=e-2$ and $\sum_{n\ge1}\frac1{(n+2)!}=e-\frac52$. Hence $S=3(e-1)-(e-2)-2\left(e-\frac52\right)=(3-1-2)e+(-3+2+5)=4$. The coefficient of $e$ vanishes, so $S$ is an integer, and a direct symbolic summation gives the same value $4$. So the series telescopes into exponential sums whose irrational parts cancel exactly, leaving the integer $S=4$ as the final numerical answer to the question. The decomposition works because $(n+2)!=(n+2)(n+1)\,n!$, so each piece reduces to a shifted factorial term, and the tail sums are standard values of the exponential series.