4
$\begingroup$

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 ?

  • 0
    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.2012-04-19
  • 0
    @joriki Thank you i updated the expression.2012-04-19
  • 2
    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.2012-04-19
  • 0
    Thanks guys i feel so foolish factorial terms should have brought binomial expansion to mind.2012-04-19

1 Answers 1

10

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 $n!$.

  • 0
    wow, thank you so much.2012-04-19
  • 1
    @Hardy: You are welcome :-)2012-04-19
  • 1
    For values other than 3, look up multisection of series.2012-04-20