Sum the series: $$(1^2+1)1! + (2^2+1)2! + (3^2+1)3! \cdots + (n^2+1)n!$$

Source: A book on algebra.I came across this interesting looking series and decided to tackle it.

My try :

All I have tried is taking the $r^{th}$ term and summing it, but in vain: $$ T_r = (r^2+1)r!$$ $$T_r = r^2\cdot r! + r!$$ Now I don't know how to sum either of these terms.I'm not familiar with uni level math as I'm a high school student. All help appreciated!


3 Answers 3


Hint. Note that $$(n^2+1)n!=n(n+1)!-(n-1)(n)!$$ Therefore the sum can be written as $$(1(2)!-(0)(1)!)+(2(3)!-(1)(2)!)+(3(4)!-(2)(3)!)+\dots +(n(n+1)!-(n-1)(n)!).$$

  • $\begingroup$ how do I go forward now? $\endgroup$ Oct 8, 2017 at 17:08
  • $\begingroup$ Should I plug in integers? $\endgroup$ Oct 8, 2017 at 17:08
  • $\begingroup$ See my next line. $\endgroup$
    – Robert Z
    Oct 8, 2017 at 17:10
  • $\begingroup$ Simple rearrangement, duh! Thank you Robert Z :) $\endgroup$ Oct 8, 2017 at 17:10
  • 1
    $\begingroup$ The right term is "telescopic sum" See en.wikipedia.org/wiki/Telescoping_series $\endgroup$
    – Robert Z
    Oct 8, 2017 at 17:12

$$ S=\sum_{r=1}^n\left( (r+2)!-3(r+1)!+2r!\right)\\ =(n+2)!+(n+1)!-3(n+1)!-3\cdot2+2\cdot1+2\cdot2\\ =n(n+1)! $$


I will prove by induction that $$\sum _{k=1}^n \left(k^2+1\right) k!=n (n+1)!\quad(*)$$ Indeed it is true for $n=1$ because $(1+1)1!=1(2!)$

Now suppose $(*)$ is true and let's prove it for $n+1$

$$\sum _{k=1}^{n+1} \left(k^2+1\right) k!=\sum _{k=1}^n \left(k^2+1\right) k!+\left[(n+1)^2+1\right](n+1)!=\\=n (n+1)!+\left[(n+1)^2+1\right](n+1)!=(n+1)!(n+n^2+2n+1+1)=\\=(n+1)!(n^2+3n+2)=(n+1)!(n+2)(n+1)=(n+1)(n+2)!$$

Hope this is useful

  • $\begingroup$ Yes, it is useful thanks @Raffaele $\endgroup$ Oct 9, 2017 at 2:25

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