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Prove that for all $n \in \mathbb{N}$,

$$n(n^2 - 1) = \frac{(n+1)!}{(n-2)!}.$$

Thanks in advance.

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Hint: $n(n^2 - 1) = n(n+1)(n-1) = (n+1)(n)(n-1)$ –  Isaac Solomon Nov 9 '13 at 16:20
    
@Pauly As you seem to be new here, I suggest you read this. –  Git Gud Nov 9 '13 at 16:21
    
Do you mean $\frac{(n+1)!}{(n-2)!}$ by any chance? –  robjohn Nov 9 '13 at 17:05
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2 Answers

up vote 7 down vote accepted

$\require{cancel}$

If you really want to express $n(n^2 - 1)$ using factorials, note that $$n(n^2 - 1) = n(n+1)(n-1) = (n+1)(n)(n-1) = \frac{(n+1)n(n-1)\cancel{(n-2)}\cdots \cancel{2}\cdot 1}{\cancel{(n - 2)}\cancel{ (n-3)}\cdots \cancel{2}\cdot 1}$$

So we have that $$n(n^2 - 1) = \dfrac{(n+1)!}{(n-2)!}$$

So the given identity you are asked to prove is incorrect: the numerator needs to be $(n + 1)!$.

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Nice guess of the actual question. –  Git Gud Nov 9 '13 at 16:34
    
Thanks, @GitGud I'm developing a knack for that! –  amWhy Nov 9 '13 at 16:35
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@GitGud I wish the "+" was not the "shift" of "=" (on the same key). I've made the same typo before :-( –  amWhy Nov 9 '13 at 16:38
    
Thank you so much! –  Pauly Nov 9 '13 at 16:39
    
You're welcome, Pauly! –  amWhy Nov 9 '13 at 16:39
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Here is how

$$n(n-1)=\frac{n(n-1)(n-2)!}{(n-2)!}=\frac{n!}{(n-2)!}$$

$$ n+1 = \frac{(n+1)!}{n!}.$$

Can you finish it?

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