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I could not compute the sum of the following series-

$$\displaystyle\sum_{k=1}^n (k^2+1)\cdot k!$$

Please tell me how to proceed in the above question. Any help is appreciated.

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  • $\begingroup$ I have tried to rearrange (k2+1) term somehow to reduce the equation but thats not working.Moreover,expandng seems to be of no help to me.Thus ,I didn't show any working as all those attempts failed. $\endgroup$
    – OtherSide0fZer0
    May 19, 2017 at 13:58

3 Answers 3

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Hint:

$$(k^2 + 1)k! = (k+2)! - 3(k+1)! + 2k!$$

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  • $\begingroup$ How did yo get that term?Please Explain. $\endgroup$
    – OtherSide0fZer0
    May 19, 2017 at 14:01
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    $\begingroup$ $k^2 + 1 = (k+2)(k+1) - 3(k+1) + 2$. $\endgroup$
    – Darth Geek
    May 19, 2017 at 14:02
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I think the answer will be - n.(n+1)! Use Mathematical Induction.

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  • $\begingroup$ This isn't true for $n=1$. $\endgroup$
    – florence
    May 19, 2017 at 14:09
  • $\begingroup$ I wrote 2 instead of 1.It was a typo.Apologies. $\endgroup$
    – BetterThanYesterday
    May 19, 2017 at 14:11
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It can be proven by induction that $$\sum_{k=1}^n (k^2+1)k! = n(n+1)!$$ This clearly holds for $n=1$. Not suppose it holds for $n$. Then $$\sum_{k=1}^{n+1}(k^2+1)k! = ((n+1)^2+1)(n+1)!+\sum_{k=1}^{n}(k^2+1)k!$$ $$=((n+1)^2+1)(n+1)! + n(n+1)!$$ $$= (n^2+3n+2)(n+1)! = (n+1)(n+2)(n+1)! = (n+1)(n+2)!$$ Thus completing the proof.

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