How do you simplify $\frac{(n + 1)!}{(n - 2)!}$ $((n+1)!)/((n-2)! )$ ?

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Answer 1 · 2018-03-02T03:00:38

$n^{3} - n$ $n^3-n$

The numerator can be rewritten as:

$\frac{(n + 1) \cdot (n + 1 - 1) \cdot (n + 1 - 2) \cdot (n + 1 - 3)!}{(n - 2)!}$ $((n+1) * (n+1-1) * (n+1-2) * (n+1-3)!)/((n-2)!)$

Simplifying:

$\frac{(n + 1) \cdot (n) \cdot (n - 1) \cdot (n - 2)!}{(n - 2)!}$ $((n+1) * (n) * (n-1) * (n-2)!) / ((n-2)!)$

We can cancel the $(n - 2)!$ $(n-2)!$ values out:

$(n + 1) (n - 1) (n)$ $(n+1)(n-1)(n)$

$= n^{3} - n$ $=n^3-n$

And there we have our answer.

How do you simplify ((n+1)!)/((n-2)! )(n+1)!(n−2)!((n+1)!)/((n-2)! )?