How to compute a simple sum

Question

In the book I am reading the author left some excersises for the reader, I happend to be stuck at this sum

$$\sum_{n=1}^k{n!(n^2+n+1)}$$

So far I have tried to factorize the polynomial, and also tried to split the sum. I know how to compute $n\ n!$, but I have no idea on the other terms, any help is accepted!

Answer 1

Note that $$\sum_{n=1}^k n!(n^2+n+1)=\sum_{n=1}^k n![(n+1)^2-n]=\sum_{n=1}^k [(n+1)!(n+1)-n!n]$$ This is a telescopic serie

Answer 2

$\textbf{Hint:}$ Write $$n^2+n+1=(n^2+3n+2)-(n+1)-(n+1)+1$$ and get a telescoping series by using $(n+1)!=(n+1)n!$