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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???

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    $\begingroup$ Hint: each term is nearly $(n+1)n!$ $\endgroup$
    – Empy2
    Commented Feb 10, 2016 at 12:29
  • $\begingroup$ 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). $\endgroup$ Commented Feb 10, 2016 at 12:41

3 Answers 3

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Hint:

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

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    $\begingroup$ Thnx, though it was the whole solution.+1 $\endgroup$ Commented Feb 10, 2016 at 12:32
  • $\begingroup$ Plz answer the edit too, thnx! $\endgroup$ Commented Feb 10, 2016 at 12:38
  • $\begingroup$ I just remembered the memory years ago that $\sum n\cdot n!$ can be represented as telescopic series. $\endgroup$ Commented Feb 10, 2016 at 12:50
  • $\begingroup$ 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. $\endgroup$
    – Dr_Be
    Commented Feb 10, 2016 at 14:16
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$$n\cdot n! = n\cdot n\cdot (n-1)! = n^2\cdot (n-1)!$$

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

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  • $\begingroup$ I got till here but after it?? $\endgroup$ Commented Feb 10, 2016 at 12:40
  • $\begingroup$ Sry bt i still am nt able to understnd, a bit more explaination plzz. $\endgroup$ Commented Feb 10, 2016 at 15:10
  • $\begingroup$ How do you get $\sum_{n = 1}^{N} n^2\cdot (n-1)! = (1 + N)! - 1$??? $\endgroup$
    – fleablood
    Commented Feb 11, 2016 at 17:11
  • $\begingroup$ @fleablood I'll write the proof later! I'm on the bus now :D $\endgroup$
    – Enrico M.
    Commented Feb 11, 2016 at 18:06
  • $\begingroup$ No, rush. (after all there are other solutions) I just think this needs clarification in your answer. $\endgroup$
    – fleablood
    Commented Feb 11, 2016 at 18:37
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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 $n=15$.

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    $\begingroup$ Sry, am a noob bt plz explain a bit more. $\endgroup$ Commented Feb 11, 2016 at 16:28
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    $\begingroup$ explain what exactly? $\endgroup$
    – Alex
    Commented Feb 11, 2016 at 16:33
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    $\begingroup$ The steps from strtng. Thnx :) $\endgroup$ Commented Feb 11, 2016 at 16:35
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    $\begingroup$ 1) Define $S_n$, 2) Add the next term on both sides, 3) Manipulate RHS with sm algebra, 4)Get the result $\endgroup$
    – Alex
    Commented Feb 11, 2016 at 16:37
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    $\begingroup$ 3rd step i dnt get it. $\endgroup$ Commented Feb 11, 2016 at 16:40

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