How do I go about summing this :
$$\sum_{r=1}^{n}r\cdot (r+1)!$$
I know how to sum up $r\cdot r!$ But I am not able to do a similar thing with this.
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2$\begingroup$ How do you sum up $r*r!$? The two sums are closely related (try breaking down $(r+1)!$ into smaller parts) $\endgroup$– TomGrubbCommented Jan 22, 2016 at 15:03
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2$\begingroup$ I helped with the formatting of your question. Please look at the modifications I made so that it will be easier in the future to use LATEX (mathJAX) formatting. There is a good guide here: meta.math.stackexchange.com/questions/5020/… $\endgroup$– TravisJCommented Jan 22, 2016 at 15:04
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$\begingroup$ @bburGsamohT by writing r as (r+1-1) then it becomes r+1! - r! In that the middle terms cancel and the answer comes out to be n+1!-1! $\endgroup$– SudhanshuCommented Jan 22, 2016 at 15:10
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$\begingroup$ I found out that the summation equals $\left(n+2\right)!-2-\sum_{r=1}^{n}\left(r+1\right)!$, but I don't think that makes things more easy. Have a look at this question for that. $\endgroup$– drhabCommented Jan 22, 2016 at 16:20
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$\begingroup$ @drhab but I think the aim of this question is to somehow manipulate it like the same way we do for r.r! $\endgroup$– SudhanshuCommented Jan 22, 2016 at 16:28
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1 Answer
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Making use of:
$$r.\left(r+1\right)!=\left(r+2-2\right).\left(r+1\right)!=\left(r+2\right)!-2.\left(r+1\right)!$$
we write the summation as:
$$\left[\left(n+2\right)!-2.\left(n+1\right)!\right]+\left[\left(n+1\right)!-2.n!\right]+\cdots+\left[4!-2.3!\right]+\left[3!-2.2!\right]$$
leading to:
$$\sum_{r=1}^{n}r.\left(r+1\right)!=\left(n+2\right)!-2-\sum_{r=1}^{n}\left(r+1\right)!$$
So finding an expression for it is in essence the same as finding an expression for: $$\sum_{r=1}^{n}r!$$
For this have a look here.
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$\begingroup$ Oh now I understood what you meant. $\endgroup$ Commented Jan 22, 2016 at 16:34