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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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    $\begingroup$ How do you sum up $r*r!$? The two sums are closely related (try breaking down $(r+1)!$ into smaller parts) $\endgroup$ – TomGrubb Jan 22 '16 at 15:03
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    $\begingroup$ 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/… $\endgroup$ – TravisJ Jan 22 '16 at 15:04
  • $\begingroup$ @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! $\endgroup$ – Sudhanshu Jan 22 '16 at 15:10
  • $\begingroup$ 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. $\endgroup$ – drhab Jan 22 '16 at 16:20
  • $\begingroup$ @drhab but I think the aim of this question is to somehow manipulate it like the same way we do for r.r! $\endgroup$ – Sudhanshu Jan 22 '16 at 16:28
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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. $\endgroup$ – Sudhanshu Jan 22 '16 at 16:34

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