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

Find
**
n
**
.

The answer is 2006.

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$\displaystyle\sum_{k=0}^n(k^2+k+1)k!=\sum_{k=0}^n((k+1)^2-k)k!=\sum_{k=0}^n\left((k+1)\cdot (k+1)!-k\cdot k!\right)$ .

This is a telescoping series, so $\displaystyle\sum_{k=0}^n(k^2+k+1)k!=(n+1)\cdot(n+1)!-0\cdot0!=(n+1)\cdot (n+1)!$ . Therefore, $(n+1)\cdot(n+1)!=2007\cdot2007!$ , so $n=\boxed{2006}$ .