# Show that $2 · 1! + 5 · 2! + 10 · 3! + . . . + (n ^2 + 1)n! = n(n + 1)!$

Show that $$2 · 1! + 5 · 2! + 10 · 3! + . . . + (n ^2 + 1)n! = n(n + 1)!$$

I got stuck after the n+1 part.

$$((n+1)^2 + 1)(n+1)! = (n+1)(n+2)!.$$

I'm not sure how to proceed from this part.

proof by Induction $$P(1):=2\cdot1!=1\cdot(1+1)!$$

Now we assume it is true for $$P(n)$$ then we proof it is true for $$P(n+1)$$, $$P(n+1):=2 · 1! + 5 · 2! + 10 · 3! + . . . + (n ^2 + 1)n! + ((n+1)^2+1)(n+1)! = (n+1)((n + 1)+1)!$$

Use the hypothesis of induction $$P(n)$$ and LHS of above expression reduces to $$n(n+1)!+((n+1)^2+1)(n+1)! = (n+1)!(n+n^2+2n+2) \\ =(n+1)!(n^2+3n+2)=(n+1)!(n+2)(n+1) \\ =(n+1)(n+2)!$$

Hence we get the RHS of the $$P(n+1)$$. So we have proved $$P(n+1)$$ is true whenever $$P(n)$$ is true.

You want to prove that $$\sum_{k=1}^n(k^2+1)k! = n(n+1)!.$$

First case: if $$n = 1$$, then $$(1^2+1)1! = 2 = 1 \cdot 2 = n(n+1)!$$.

Then by induction, if the formula holds for $$n-1$$ you find: $$\sum_{k=1}^n(k^2+1)k! = \sum_{k=1}^{n-1}(k^2+1)k! + (n^2+1)n! = (n-1)(n-1+1)! + (n^2+1)n! = (n-1)n! + (n^2+1)n! = (n+n^2)n! = n(n+1)n! = n(n+1)!$$

Alternatively: $$2 · 1! + 5 · 2! + 10 · 3! + . . . + (n ^2 + 1)n! =\sum_{i=1}^n(i^2+1)i!=\\ \sum_{i=1}^n[(i+1)^2-i-i]i!=\\ \color{red}{\sum_{i=1}^n(i+1)\cdot (i+1)!-\sum_{i=1}^n i\cdot i!}-\color{blue}{\sum_{i=1}^n (i+1-1)\cdot i!}=\\ \color{red}{(n+1)(n+1)!-1}-\color{blue}{(n+1)!+1}= n(n + 1)!$$