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 ?

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 $n!$.
