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$$1 + \frac{1^2 + 2^2}{2!} + \frac{{1}^2 + {2}^2 + 3^2}{3!} + \cdots$$

I can't figure out how to do summations which involve a factorial term in the denominator. Please help. This is a past year IITJEE question by the way.

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  • $\begingroup$ Don't expect a trivial solution to this... $\endgroup$
    – Zach466920
    Apr 10, 2015 at 18:44
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    $\begingroup$ @Zach466920 Not trivial, perhaps, but definitely standard and leading to $$\frac{17e}6.$$ $\endgroup$
    – Did
    Apr 10, 2015 at 18:56
  • $\begingroup$ @Did Your right, I didn't think to sum the integers and then sum the terms. Still difficult, but yep now it looks pretty trivial. $\endgroup$
    – Zach466920
    Apr 10, 2015 at 19:07
  • $\begingroup$ See Faulhaber's formulas. $\endgroup$
    – Lucian
    Apr 10, 2015 at 19:36

1 Answer 1

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HINT:

$$\sum_{r=1}^n\dfrac{1^2+2^2+\cdots+r^2}{r!}=\frac16\sum_{r=1}^n\dfrac{r(r+1)(2r+1)}{r!}=\frac16\sum_{r=1}^n\dfrac{r(r+1)(2r+1)}{r!}$$

Now for $r>0,$ $$\dfrac{r(r+1)(2r+1)}{r!}=\dfrac{(r+1)(2r+1)}{(r-1)!}$$

Let $(r+1)(2r+1)=2(r-1)(r-2)+a(r-1)$

Set $r=2$ to get $a$

Now, $$e^x=\sum_{r=0}^\infty\dfrac{x^u}{u!}$$

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