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In a contest between me and my friend, i was able to solve all the questions till he stumped me at this one.

$ \sum_{n=1}^{15}n(n!) =?$

The only thing I could think of how to pursue is

$n(n+1)! = n^2(n-1)!$.

Preferably give hints, so i can try it again.

Edit: After getting the solution the next second I posted the question, I would further extend the question to how you guys got the idea, of this fairly simple looking but well thought out solution. Is it all practice or you might get some hint always from the question???

Enrico M.
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Display name
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    Hint: each term is nearly $(n+1)n!$ – Empy2 Feb 10 '16 at 12:29
  • You can rephrase the 'Edit' section to "how does the human mind works?". I don't think you'll get a good enough answer here (or anywhere else for that matter, unless you can time-travel to the future). – barak manos Feb 10 '16 at 12:41

3 Answers3

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Hint:

$$ n\cdot n! = (n+1-1)n!=(n+1)n!-n! = (n+1)!-n! $$

choco_addicted
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    Thnx, though it was the whole solution.+1 – Display name Feb 10 '16 at 12:32
  • Plz answer the edit too, thnx! – Display name Feb 10 '16 at 12:38
  • I just remembered the memory years ago that $\sum n\cdot n!$ can be represented as telescopic series. – choco_addicted Feb 10 '16 at 12:50
  • This reminds me of an exercise back at the University which was similar and involved a telescopic series, too. I was so puzzled until I saw the solution. My recommendation to the OP: Just try to memorize it and use it in the right places. No worries, this is the normal process of learning and getting more and more experienced. – Dr_Be Feb 10 '16 at 14:16
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$$n\cdot n! = n\cdot n\cdot (n-1)! = n^2\cdot (n-1)!$$

$$\sum_{n = 1}^{N} n\cdot n! = (1 + N)! - 1 $$

Enrico M.
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Use perturbation method from 'Concrete Mathematics': $$ S_n = \sum_{k=1}^{n} k! \to S_n + (n+1)! = \sum_{k=1}^{n} k! + (n+1)! = \sum_{k=1}^{n}(k+1)! + 1 = \sum_{k=1}^{n} k!k + \sum_{k=1}^{n} k! + 1 $$ Now do some very simple algebra and get the result. Then plug in your $n=15$.

Alex
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    3rd step i dnt get it. – Display name Feb 11 '16 at 16:40
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    $1+ 2! + \ldots n! + (n+1)! = \sum_{k=1}^{n} (k+1)! + 1$ – Alex Feb 11 '16 at 16:42
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    Thnx n now the last step. – Display name Feb 11 '16 at 16:45
  • Use the property of factorial and cancel out the $S_n$ term – Alex Feb 11 '16 at 16:46
  • Uhm, sorry but this does not strike me as a good answer. It is a little involved, the first steps being completely useless: the point of the question is just the last step, which is not clearly motivated. – bartgol Feb 11 '16 at 16:56
  • I'm practically devastated – Alex Feb 11 '16 at 17:00
  • This does seem a bit obscure and hiding a really good answer. I think it'd be more direct and clearer as: $\sum_{k=1}^{n+1} k! = \sum_{k=0}^n(k+1)! =\sum_{k=0}^nk!(k+1) = \sum_{k=0}^n(k!k+k!) = \sum_{k=0}^nk!k + \sum_{k=0}^nk! = \sum_{k=1}^nk!k + \sum_{k=1}^nk! + 1 \implies \sum_{k=1}^nk!k = \sum_{k=1}^{n+1} k! - \sum_{k=1}^nk! - 1 = (n+1)! - 1$. – fleablood Feb 11 '16 at 18:57
  • This approach is from Concrete Mathematics and is totally legitimate – Alex Feb 16 '16 at 08:23