Sum of the series ∑r=1n r2+1(r!) is

We can writer2+1=(r+2)(r+1)−3(r+1)+2Thus ∑r=1n r2+1(r!)=∑r=1n [(r+2)(r+1)−(r+1)−2{(r+1)−1}]r! =∑r=1n [(r+2)!−(r+1)!]−2∑r=1n {(r+1)!−r!}=(n+2)!−2!−2{(n+1)!−1}=n(n+1)!