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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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  • $\begingroup$ First you add $ (n+1)!+(n+1)! $ together? $\endgroup$ – user23053 Jan 23 '12 at 14:41
  • $\begingroup$ @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. $\endgroup$ – Gigili Jan 23 '12 at 15:13
  • $\begingroup$ Hope you don't mind my formatting. $\endgroup$ – Rasmus Jan 23 '12 at 15:27
  • $\begingroup$ Where are you? +1 $\endgroup$ – mrs Nov 26 '12 at 15:14
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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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  • $\begingroup$ +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. $\endgroup$ – 000 May 19 '12 at 6:30

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