<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] 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).
$$= 100! \times (1 + [-\frac{1}{0!} + \frac{101}{100!}])$$
$$= 100! \times (1 - 1 + \frac{101}{100!})$$
$$= 100! \times \frac{101}{100!}$$
$$= 101$$
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