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I haven't had to prove something of this type before, and can't even figure out how to start a relevant proof. Induction has been the general go-to method, but the switch (and combination) from i to n is somehow messing with my ability to get anywhere.

Any help would be appreciated.

3 Answers 3

7

$$\sum_{i=0}^n (i+1) \times (i+1)!= \sum_{i=1}^{n+1} i \times i!= \sum_{i=1}^{n+1} (i+1-1) \times i!= \sum_{i=1}^{n+1} (i+1)!-i!$$ Now , the terms cancel each other, leaving $(n+2)!-1$ as desired.

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    excellent (+1)! – 2017-01-25
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    :D Basically what I was aiming for! – 2017-01-25
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    Thank you both. I guess I just don't have enough experience with summation to make the intuitive leap to that first step (or any after), nothing like that even occurred to me. – 2017-01-25
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    @Bob :D Finding the sneaky telescoping series often comes with practice ;). 3 months ago I probably wouldn't have seen it either. – 2017-01-25
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$$(i+1)(i+1)!=(i+2)(i+1)!-(i+1)!=(i+2)!-(i+1)!$$

So you get an alternating series:

$$\sum_{i=0}^n(i+1)(i+1)!=\sum_{i=0}^n(i+2)!-(i+1)!=(n+2)!-(0+1)!$$

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    also nice (+1)! – 2017-01-25
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    @Simply Beautiful Art: Can you explain how you went (or what made you think to go) from $(i+1)(i+1)!$ to $(i+2)(i+1)! - (i+1)!$? – 2017-01-25
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    @Bob If you factor out the $(i+1)!$, what do you get? – 2017-01-25
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    $(i+1)!((i+2)-1) = (i+1)!(i+1)$ Damn, is that supposed to be obvious? Because I completely (obviously) missed it. Not the solution to factoring it out, but to go from the first to the second? – 2017-01-25
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    @Bob Well, it's a factorial. So... practice... let's you see that. – 2017-01-25
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Some of the answers here are quite elegant. But since it is not here I will show you a proof by induction. It is a bit of a hammer, but it a hammer is a useful tool.

base case:

$n = 0\\ 1\cdot 1! = 1 = 2! - 1$

Inductive hypothesis:

Assume $\sum_\limits{i=0}^n (i+1)(i+1)! = (n+2)! -1.$

We will show that $\sum_\limits{i=0}^{n+1} (i+1)(i+1)! = (n+3)! -1,$ based on the inductive hypothesis.

$\sum_\limits{i=0}^{n+1} (i+1)(i+1)! = $$\sum_\limits{i=0}^{n} (i+1)(i+1)! + (n+2)(n+2)!\\(n+2)! - 1 + (n+2)(n+2)!\\ (n+3)(n+2)! - 1\\ (n+3)! - 1$

QED