Convert $\sum_{k=1}^n k^2(k+1)!$ to closed form

Question

Convert the following to closed form. $$\sum_{k=1}^n k^2(k+1)!$$

I've been trying to solve this .. no luck. You don't have to solve the whole thing, just point me in the right direction. Thank you!

Answer 1

Write $k^2$ in terms of $k+ 3$ and $k+ 2$ so that you will only have factorials, $$k^2= (k+3)(k+2)- 5(k+ 2)+ 4.$$ Then note that the sum of factorials cancel leaving only $$(n+ 3)!- 4(n+ 2)!+ 2$$

Answer 2

Warm up:

We have

$$(k+1)!-k!=kk!$$ so that the sum of $kk!$ telescopes to $$(n+1)!-1.$$

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Main course:

We have $$(k+2)!=(k+2)(k+1)!,\\(k+3)!=(k^2+5k+6)(k+1)!,$$

then

$$(k+3)!-5(k+2)!+4(k+1)!=k^2(k+1)!.$$

But

$$(k+3)!-5(k+2)!+4(k+1)!=((k+3)!-(k+2)!)-4((k+2)!-(k+1)!)$$ and we have telescoping sums giving

$$((n+3)!-2)-4((n+2)!-1).$$