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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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    Thanks guys i feel so foolish factorial terms should have brought binomial expansion to mind.2012-04-19

1 Answers 1

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Start with

$ (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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    For values other than 3, look up multisection of series.2012-04-20