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

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Answer 1 · 2018-03-02T15:11:43

$n^{2} + 3 n + 2$ $n^2+3n+2$

We can rewrite the numerator as:

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

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

We can cancel $(n)!$ $(n)!$ and $(n)!$ $(n)!$ out:

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

$= (n + 2) \cdot (n + 1)$ $=(n+2) * (n+1)$

$= n^{2} + 3 n + 2$ $=n^2+3n+2$

Thus, solved

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