# Multiply Factorial

I have a small comprehension gap with an easy equation. I have following term and I don’t know how to multiply it correctly. $(n+1)(n+1)!+(n+1)!-1$

One intermediate step must be. $(n+2)(n+1)!-1$

The result should be $(n+2)!-1$.

How do I multiply the term correct? Is the attempt correct to multiply binomial series to $(n+1)^2n!+(n+1)!-1$?

It would awesome, if someone could help me.

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$$(n+1)\color{red}{(n+1)!}+\color{red}{(n+1)!}-1$$ $$=(n+1)![(n+1)+1]-1$$ $$=(n+2)(n+1)!-1$$ $$=(n+2)!-1$$

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First you add $(n+1)!+(n+1)!$ together? –  hofmeister Jan 23 '12 at 14:41
@Taz: No, I factored $(n+1)!$ first, which is common as I showed with red. What remains from first statement is $(n+1)$ and $1$ from the second one, which will be $(n+2)$. Multiplying $(n+2)$ and $(n+1)!$, we get $(n+2)!$. I hope it's clear now. –  Gigili Jan 23 '12 at 15:13
Thanks now i got it! –  hofmeister Jan 23 '12 at 15:24
Hope you don't mind my formatting. –  Rasmus Jan 23 '12 at 15:27
Where are you? +1 –  Babak S. Nov 26 '12 at 15:14

You would have no trouble with $(5)(n+1)!+(7)(n+1)!=(12)(n+1)!$ ($5$ apples plus $7$ apples equals $12$ apples.)

Maybe you would have a little trouble with $(5)(n+1)!+(n+1)!$, but not if you rewrite it as $(5)(n+1)! +(1)(n+1)!$ ($5$ apples plus $1$ apple equals $6$ apples).

Now let's look at $(n+1)(n+1)! +(n+1)!$. Rewrite this as $(n+1)(n+1)! +(1)(n+1)!$. We have $n+1$ apples plus $1$ apple is $n+2$ apples, so the sum is $(n+2)(n+1)!$, which can be rewritten as $(n+2)!$.

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+1 for two reasons: absurdly and casually turning the factorial into an apple (there is academic merit in that) and the coincidence I'm eating two apples. –  000 May 19 '12 at 6:30