求和：$\dfrac{3}{1 !+2 !+3 !}+\dfrac{4}{2 !+3 !+4 !}+\cdots+\dfrac{n+2}{n !+(n+1) !+(n+2) !}$
答案 $\dfrac12-\dfrac{1}{(n+2)!}$．
解析 设题中代数式为 $m$，则\[\begin{split} m&=\sum_{k=1}^n\dfrac{k+2}{k!+(k+1)!+(k+2)!}\\ &=\sum_{k=1}^n\dfrac{k+2}{k!\big(1+k+1+(k+2)(k+1)\big)}\\ &=\sum_{k=1}^n\dfrac{1}{(k+2)\cdot k!}\\ &=\sum_{k=1}^n\dfrac{k+1}{(k+2)!}\\ &=\sum_{k=1}^n\left(\dfrac{1}{(k+1)!}-\dfrac{1}{(k+2)!}\right)\\ &=\dfrac12-\dfrac{1}{(n+2)!},\end{split}\]