<s>[INST] Exact value of $100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)$ Problem
Find the exact value of $$100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)$$
What I tried :
multiply $(n-1)$ to sumnation's numerator and denominator then it changed to 
$$100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^3-1)}{n(n-2)!}\right)$$
But this didn't give me any clues to solve this.
I just guess telescoping is the way to solve because this is finite-sum, but I don't have any idea.
 [/INST] Want
$100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)
$
Going up to $m$,
$\begin{array}\\
\sum_{n=1}^{m} \dfrac{(-1)^n (n^2+n+1)}{n!}
&=\sum_{n=1}^{m} \dfrac{(-1)^n (n^2-n+2n+1)}{n!}\\
&=\sum_{n=1}^{m} \dfrac{(-1)^n (n^2-n+2n+1)}{n!}\\
&=\sum_{n=1}^{m} (-1)^n(\dfrac{ n^2-n}{n!}+\dfrac{2n}{n!}+\dfrac{ 1}{n!})\\
&=\sum_{n=1}^{m} (-1)^n\dfrac{ n^2-n}{n!}+2\sum_{n=1}^{m} (-1)^n\dfrac{ n}{n!}+\sum_{n=1}^{m} (-1)^n\dfrac{ 1}{n!}\\
&=\sum_{n=2}^{m} (-1)^n\dfrac{ n(n-1)}{n!}+2\sum_{n=1}^{m} (-1)^n\dfrac{ 1}{(n-1)!}+\sum_{n=1}^{m} (-1)^n\dfrac{ 1}{n!}\\
&=\sum_{n=2}^{m} (-1)^n\dfrac{ 1}{(n-2)!}+2\sum_{n=0}^{m-1} (-1)^{n+1}\dfrac{ 1}{n!}+\sum_{n=1}^{m} (-1)^n\dfrac{ 1}{n!}\\
&=\sum_{n=0}^{m-2} (-1)^n\dfrac{ 1}{n!}-2\sum_{n=0}^{m-1} (-1)^{n}\dfrac{ 1}{n!}+\sum_{n=1}^{m} (-1)^n\dfrac{ 1}{n!}\\
&=1+\sum_{n=1}^{m-2} (-1)^n\dfrac{ 1}{n!}-2-2\sum_{n=1}^{m-2} (-1)^{n}\dfrac{ 1}{n!}-2\dfrac{(-1)^{m-1}}{(m-1)!}\\
&\quad +\sum_{n=1}^{m-2} (-1)^n\dfrac{ 1}{n!}+\dfrac{(-1)^{m-1}}{(m-1)!}+\dfrac{(-1)^m}{m!}\\
&=-1-2\dfrac{(-1)^{m-1}}{(m-1)!}+\dfrac{(-1)^{m-1}}{(m-1)!}+\dfrac{(-1)^m}{m!}\\
&=-1+\dfrac{(-1)^{m}}{(m-1)!}+\dfrac{(-1)^m}{m!}\\
\end{array}
$
so
$\begin{array}\\
m!\left(1+\sum_{n=1}^{m} \frac{(-1)^n (n^2+n+1)}{n!}\right)
&=m!(1+(-1+\dfrac{(-1)^{m}}{(m-1)!}+\dfrac{(-1)^m}{m!})\\
&=(-1)^{m}m+(-1)^m)\\
&=(-1)^{m}(m+1)\\
\end{array}
$
For $m=100$,
this is
$101$.
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