Evaluate $\sum_{k=1}^{n}(k^2 \cdot (k+1)!)$ We have to evaluate the following:
$$1^2 \cdot 2! + 2^2 \cdot 3! + \cdots + n^2 \cdot (n+1)! =\sum_{k=1}^{n} k^2 \cdot (k+1)!$$
Any hints ?

Using k2=(k+3)(k+2)−5(k+2)+4 we write
fk=k2(k+1)!=(k+3)!−5(k+2)!+4(k+1)!=Gk+1−Gk
where Gk=(k+2)!−4(k+1)!, therefore
∑k=1nfk=∑k=1n(Gk+1−Gk)=Gn+1−G1
Since G1=−2, and Gn+1=(n+3)!−4(n+2)!=(n+2)!(n−1) we get
∑k=1nk2(k+1)!=(n−1)(n+2)!+2
