5
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Evaluate for a fixed $m\neq 1$ ( $m\in \mathbb{N}$ )

$$\sum _{k=1}^{n}\left[\left( \sum _{i=1}^{k}i^{2}\right) \left(\sum _{k_{1}+k_{2}+...+k_{m}=k}\dfrac {\left( k_{1}+k_{2}+\ldots +k_{m}\right) !} {k_{1}!k_{2}!...k_{m}!}\right)\right]$$

  • 0
    Note that the last part of your summation: $$\sum_{k_{1}+k_{2}+\cdots+k_{m}=k}{\frac{k!}{k_{1}!k_{2}!\cdots k_{m}!}}=\sum_{k_{1}+k_{2}+\cdots+k_{m}=k}{{k \choose k_{1},k_{2},\cdots,k_{m}}}$$ Where ${k \choose k_{1},k_{2},\cdots,k_{m}}$ is the [multinomial co-efficient](http://math.stackexchange.com/a/151615/32374).2012-07-23
  • 1
    Is your innermost sum (1) over all $m$-vectors of natural numbers whose sum of components is $k$, (2) over all $m$-vectors of positive integers whose sum of components is $k$, (3) over all $m$-component partitions of $k$, or (4) something else?2012-07-23

2 Answers 2