$\sum_{k=1}^n(k!)(k^2+k+1)$ for $n=1,2,3...$ and obtain an expression in terms of $n$

Question

Find a closed expression in terms of $n$. $$\sum_{k=1}^n(k!)(k^2+k+1); n=1,2,3...$$
Any idea about how to do this.. I'm a new to this so a little explanation would be helpful. Thanks in advance!

Answer 1

Write $k^2+k+1 = (k+1)^2-k$ so that

$$(k!)(k^2+k+1) = (k+1)(k+1)! - k \, k!$$

Now you have a telescoping sum.

Answer 2

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

This is the req. sum.