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I don't understand how $$ \frac {n(n-1)!+(n+1)n(n-1)!}{n(n-1)!-(n-1)!} $$

becomes $$ \frac {(n-1)![n+(n+1)n]}{(n-1)!(n-1)} $$

and then how it becomes $$ \frac {n+n^2+n}{n-1} $$

I've tried applying the distributive property but I seem to be missing something.

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    $\begingroup$ In the first step you factored out a common factor of $(n-1)!$ in both the numerator and the denominator. In the second step you cancelled the common factor you took out in the previous step. Also in the second step, on top you multiplied out $(n+1)n=n^2+n$. $\endgroup$ – paw88789 Oct 25 '14 at 22:35
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Think of it like this:

$$\frac {n(n-1)!+(n+1)n(n-1)!}{n(n-1)!-(n-1)!}$$

Define $$D:=(n-1)!$$ then

$$\frac {n(n-1)!+(n+1)n(n-1)!}{n(n-1)!-(n-1)!}=\frac {nD+(n+1)nD}{nD-D}$$ now factor $D$

$$\frac {(n+(n+1)n)D}{(n-1)D}$$ now the D's cancel, $$\frac {(n+(n+1)n)}{(n-1)}$$ $$\frac {(n+n^2+n)}{(n-1)}$$ $$\frac {n+n^2+n}{n-1}$$ $$\frac {n^2+2n}{n-1}$$

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  • $\begingroup$ It makes sense now, thanks! $\endgroup$ – Jet_Set_Willy Oct 26 '14 at 0:11

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