# Summing an unusual series: $\frac {x} {2!(n-2)!}+\frac {x^{2}} {5!(n-5)!}+\dots +\frac {x^{\frac{n}{3}}} {(n-1)!}$

How to sum the following series $$\frac {x} {2!(n-2)!}+\frac {x^{2}} {5!(n-5)!}+\frac {x^{3}} {8!(n-8)!}+\dots +\frac {x^{\frac{n}{3}}} {(n-1)!}$$ n being a multiple of 3.

This question is from a book, i did not make this up. I can see a pattern in each term as the ith term can be written as $\frac {x^i}{(3i-1)!(n+1-3i)!}$

but i am unsure what it going on with the indexing variable's range. Any help would be much appreciated ?

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Your pattern contains unbalanced parentheses; it's probably missing a closing parenthesis and a factorial? Also I strongly suspect that you want $2!(n-2)!$ etc. in the denominators. Note that the factorial takes precedence over arithmetic operators. –  joriki Apr 19 '12 at 23:10
@joriki Thank you i updated the expression. –  Hardy Apr 19 '12 at 23:16
The terms are probably of the shape $\frac{1}{k!(b-k)!}$, with some power of $x$. Multiply by $b!$ so they will look like binomial coefficients. So the series looks like a binomial expansion with some missing terms. (You will want to let $x=y^3$.) Then use cube roots of unity in binomial expansion to kill two thirds of the terms. –  André Nicolas Apr 19 '12 at 23:22
Thanks guys i feel so foolish factorial terms should have brought binomial expansion to mind. –  Hardy Apr 19 '12 at 23:32

$$(1 + x)^n = \sum_{r=0}^{n} \binom{n}{r} x^r$$

Multiply by $x$

$$f(x) = x(1 + x)^n = \sum_{r=0}^{n} \binom{n}{r} x^{r+1}$$

Now if $w$ is a primitive cube-root of unity then

$$f(x) + f(wx) + f(w^2 x) = 3\sum_{k=1}^{n/3} \binom{n}{3k-1} x^{3k}$$

Replace $x$ by $\sqrt[3]{x}$ and divide by $n!$.

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wow, thank you so much. –  Hardy Apr 19 '12 at 23:27
@Hardy: You are welcome :-) –  Aryabhata Apr 19 '12 at 23:31
For values other than 3, look up multisection of series. –  marty cohen Apr 20 '12 at 2:14