TITLE: How to find the sum $\sum\limits_{k=1}^n (k^2+k+1)k!$?
QUESTION [6 upvotes]: How to find the sum $$\sum^n_{k=1} (k^2+k+1)k!$$
What I tried as follows : 
$$\sum^n_{k=1} (k^2+k+1)k!$$
=$$\sum^n_{k=1} (k^2)k!+ \sum^n_{k=1} (k)k! + \sum^n_{k=1} (1)k!$$
Now we can write $\sum^n_{k=1} (k^2) $ as $$\frac{n(n+1)(2n+1)}{6}$$ but what to do with $k!$ please guide thanks.......

REPLY [8 votes]: Given expression is the same as $$\sum_{k=1}^n ((k+1)^2-k)k!=\sum_{k=1}^n(k+1)(k+1)!-\sum_{k=1}^nkk!\\
=\sum_{k=2}^{n+1}kk!-\sum_{k=1}^nkk!=(n+1)(n+1)!-1!1=(n+1)!(n+1)-1$$