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.
 A: We have
$$100! \times (1 + \sum_{n=1}^{100} \frac{(-1)^n(n^2 + n + 1)}{n!})$$
$$= 100! \times (1 + \sum_{n=1}^{100} (-1)^n (\frac{n}{(n-1)!} + \frac{n+1}{n!}))$$
because $\frac{n^2 + n + 1}{n!} = \frac{n}{(n-1)!} + \frac{n+1}{n!}$. Expanding, we have
$$= 100! \times (1 + [-\frac{1}{(1-1)!} - \frac{1+1}{1!} + \frac{2}{(2-1)!} + \frac{2+1}{2!} - \dots - \frac{99}{(99-1)!} - \frac{99+1}{99!} + \frac{100}{(100-1)!} + \frac{100 + 1}{100!}])$$
You should be able to see that all terms are cancelled out apart from the first and the last (telescoping, as you suspected).
$$= 100! \times (1 + [-\frac{1}{0!} + \frac{101}{100!}])$$
$$= 100! \times (1 - 1 + \frac{101}{100!})$$
$$= 100! \times \frac{101}{100!}$$
$$= 101$$
A: Write the sum like
$$\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}+\frac{(-1)^n n}{n!} + \frac{(-1)^n}{n!}$$
$$=\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}+\sum_{n=1}^{100}\frac{(-1)^n n}{n!} +\sum_{n=1}^{100} \frac{(-1)^n}{n!}$$
$$=\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}+\sum_{n=0}^{99}\frac{(-1)^{n+1}}{n!} +\sum_{n=1}^{100} \frac{(-1)^n}{n!}$$
$$=\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}+\sum_{n=0}^{99}\frac{(-1)^{n+1}}{n!} +\sum_{n=1}^{100} \frac{(-1)^n}{n!}$$
$$=\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}-1+\frac{1}{100!}.$$
Because only the zeroth term of the second sum and the 100th term of the third sum survive.  Now your expression is
$$100! \times \left(1+\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}-1+\frac{1}{100!}\right)$$
$$= 100!\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!} +1.$$
If $n$ is odd and at least three, then two successive terms of the remaining sum are
$$\frac{-n}{(n-1)!}+\frac{n+1}{n!} = \frac{-1}{(n-2)!}+\frac{1}{n!}.$$  So the sum from 3 to 100 telescopes to $-1 + 1/99!.$  Tacking on the $n=1$ and $n=2$ terms gives us
$$100!\left(\frac{-1}{0!} +\frac{2}{1!} -1 +\frac{1}{99!}\right) +1 = 101.$$
A: $$
\begin{align}
&100!\,\left(1+\sum_{n=1}^{100}\frac{(-1)^n(n^2+n+1)}{n!}\right)\tag1\\
&=100!\,\left(\sum_{n=0}^{100}\frac{(-1)^n(n^2+n+1)}{n!}\right)\tag2\\
&=100!\,\left(\sum_{n=0}^{100}\frac{(-1)^n(\color{#C00}{n(n-1)}\color{#090}{+2n}\color{#00F}{+1})}{n!}\right)\tag3\\
&=100!\,\left(\color{#C00}{\sum_{n=0}^{98}\frac{(-1)^n}{n!}}\color{#090}{-2\sum_{n=0}^{99}\frac{(-1)^n}{n!}}\color{#00F}{+\sum_{n=0}^{100}\frac{(-1)^n}{n!}}\right)\tag4\\
&=100!\,\left(\color{#090}{\frac2{99!}}+\color{#00F}{\frac1{100!}-\frac1{99!}}\right)\tag5\\[9pt]
&=101\tag6
\end{align}
$$
Explanation:
$(2)$: the $n=0$ term is $1$
$(3)$: rewrite $n^2+n+1$ as $n(n-1)+2n+1$
$(4)$: substitute $\color{#C00}{n\mapsto n+2}$ and $\color{#090}{n\mapsto n+1}$
$(5)$: everything for $n=0$ to $n=98$ cancels; what's left is here
$(6)$: simplify
A: 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
$101$.
