Sum $\sum\limits_{k=1}^n (k^2+1)\cdot k!$

Question

I could not compute the sum of the following series-

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

Please tell me how to proceed in the above question. Any help is appreciated.

Answer 1

I think the answer will be - n.(n+1)! Use Mathematical Induction.

Answer 2

Hint:

$$(k^2 + 1)k! = (k+2)! - 3(k+1)! + 2k!$$

Answer 3

It can be proven by induction that $$\sum_{k=1}^n (k^2+1)k! = n(n+1)!$$ This clearly holds for $n=1$. Not suppose it holds for $n$. Then $$\sum_{k=1}^{n+1}(k^2+1)k! = ((n+1)^2+1)(n+1)!+\sum_{k=1}^{n}(k^2+1)k!$$ $$=((n+1)^2+1)(n+1)! + n(n+1)!$$ $$= (n^2+3n+2)(n+1)! = (n+1)(n+2)(n+1)! = (n+1)(n+2)!$$ Thus completing the proof.