# Simplifying a Product of Summations

I have, for a fixed and positive even integer $n$, the following product of summations:

$\left ( \sum_{i = n-1}^{n-1}i \right )\cdot \left ( \sum_{i = n-3}^{n-1} i \right )\cdot \left ( \sum_{i = n-5}^{n-1}i \right )\cdot ... \cdot \left (\sum_{i=5}^{n-1}i \right )\cdot \left (\sum_{i=3}^{n-1}i \right )\cdot \left (\sum_{i=1}^{n-1}i \right )$

Where there are $\frac{n}{2}$ groups of summations multiplied together.

For example, consider the case where $n=4$ :

$\left ( \sum_{i = 3}^{3}i \right )\cdot \left ( \sum_{i = 1}^{3} i \right ) = \left ( 3 \right )\left ( 1+2+3 \right ) = 18$

I have tried in vain to simplify the product. Perhaps there are identities I could make use of.

Edit : I can expand the product to clarify:

$$\left ( n-1 \right )\cdot \left [ (n-3)+(n-2)+(n-1) \right ]\cdot \left [ (n-5)+...+(n-1) \right ]\cdot ... \cdot\left [3+4+...+(n-1)\right ]\cdot \left [1+2+...+(n-1) \right ]$$

From where I can see a $(n-1)^{\frac{n}{2}}$ term, but the others are quite jumbled.

• I don't see any particular way this is going to simplify. I would actually work out the values of each sum (they're just arithmetic progressions) - for example, the last factor is just $(n-1)n/2$ - that should make things a little simpler, once the sums are gone. – Greg Martin May 23 '13 at 18:25

Let $n=2m$. Your expression is then
\begin{align*} &(2m-1)\cdot\frac32(2n-4)\cdot\frac52(2n-6)\cdot\ldots\cdot\frac{n-1}2n\\\\ &\qquad=1\cdot(2m-1)\cdot3(2m-2)\cdot5(2m-3)\cdot\ldots\cdot(2m-1)m\\\\ &\qquad=\prod_{k=1}^m(2k-1)(2m-k)\\\\ &\qquad=(2m-1)!!\frac{(2m-1)!}{(m-1)!}\\\\ &\qquad=\frac{(2m)!}{2^mm!}\cdot\frac{(2m-1)!}{(m-1)!}\\\\ &\qquad=\frac{(2m)!}{2^m}\binom{2m-1}m\\\\ &\qquad=\frac{n!}{2^{n/2}}\binom{n-1}{n/2}\;. \end{align*}
Note that $\displaystyle \sum_{i=n - 2k +1}^{n-1} i = (2k - 1)(n-k)$. Let $n = 2m$ \begin{align*}\prod_{k = 1}^{\frac{n}{2}} (2k - 1)(n- k) &= \prod_{k = 1}^{m} (2k - 1) \prod_{k=1}^m (2m - k) \\ &=(2m-1)!! \frac{(2m-1)!}{(m-1)!} \end{align*}