Find the value of n if:

Question

$$\sum_{k=0}^n (k^{2}+k+1) k! = (2007).2007!$$

How to approach this problem? In need of ideas. Thank you.

Answer 1

Note that $$ \begin{align} (n+1)(n+1)!-n\,n! &=\left((n+1)^2-n\right)n!\\ &=\left(n^2+n+1\right)n! \end{align} $$ Therefore, $$ \begin{align} \sum_{n=0}^{2006}\left(n^2+n+1\right)n! &=\sum_{n=0}^{2006}\left((n+1)(n+1)!-n\,n!\right)\\ &=\sum_{n=1}^{2007}n\,n!-\sum_{n=0}^{2006}n\,n!\\[9pt] &=2007\cdot2007!-0\cdot0! \end{align} $$