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Find the sum of the following series: $$2 \cdot 1! + 5 \cdot 2! + 10 \cdot 3! + 17 \cdot 4! + \cdots + (n^2 +1)n!$$

Since the question is asking about the closed form of its sum, I thought it must be some telescoping series. So I tried to express it in the form of $f(n+1)-f(n)$, but my attempt turned out to be futile, below is my attempt on the question:

since $n^2 + 1 = (n+1)^2-2n$,

$$(n^2 +1)n! = n![(n+1)^2-2n] = n!(n+1)^2-n!2n$$

But it isn't in the form of the telescoping series when I expressed it, because

for $n$ we have $$n!(n+1)^2-n!2n$$

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

The terms don't cancel each other... can someone please guide me through the question? perhaps some hints? thanks in advance

Michael Hardy
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Dave Clifford
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  • Please use MathJax, I can't read your equations; I don't even know what your series looks like. Also, factorial for rational numbers seems an unlikely expression here ($2.1! = \frac{21}{10}!$ ?!?) – AlexR Sep 11 '13 at 11:38
  • sorry but I don't know how to use mathjax, the dot in the series represents multiply, 2.1! = 2x1! , 5.2! = 5x2! and so on – Dave Clifford Sep 11 '13 at 11:39
  • I did it for you now, please take a look at the code (just "edit" and see the $'es). It's not that hard and there are many resources (Look up the Math.SE faq or any introduction to (La)TeX. – AlexR Sep 11 '13 at 11:43
  • $n^2+1 = (n+1)^2-2n$, not $(n^2+1)-2n$ – roman Sep 11 '13 at 11:43
  • @AlexR Thanks for your generous help! I will study how to use MathJax ASAP – Dave Clifford Sep 11 '13 at 11:44
  • @roman that was a typing error, thanks for pointing it out – Dave Clifford Sep 11 '13 at 11:45

1 Answers1

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$$ (k^2+1) \cdot k! = (k+1) \cdot (k+1)! - 2k \cdot k! = \left((k+2)! - (k+1)!\right) - 2\left((k+1)! - k!\right) $$

easymath3
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