Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
add comment

2 Answers 2

up vote 5 down vote accepted

Let $n=2m$. Your expression is then

$$\begin{align*} &(2m-1)\cdot\frac32(2n-4)\cdot\frac52(2n-6)\cdot\ldots\cdot\frac{n-1}2n\\\\ &\qquad=1\cdot(2m-1)\cdot3(2m-2)\cdot5(2m-3)\cdot\ldots\cdot(2m-1)m\\\\ &\qquad=\prod_{k=1}^m(2k-1)(2m-k)\\\\ &\qquad=(2m-1)!!\frac{(2m-1)!}{(m-1)!}\\\\ &\qquad=\frac{(2m)!}{2^mm!}\cdot\frac{(2m-1)!}{(m-1)!}\\\\ &\qquad=\frac{(2m)!}{2^m}\binom{2m-1}m\\\\ &\qquad=\frac{n!}{2^{n/2}}\binom{n-1}{n/2}\;. \end{align*}$$

share|improve this answer
    
This is exactly what I was looking for. Thanks a lot! –  FlamingWilderbeest May 23 '13 at 20:00
    
@FlamingWilderbeest: You’re welcome! –  Brian M. Scott May 23 '13 at 20:14
add comment

Note that $ \displaystyle \sum_{i=n - 2k +1}^{n-1} i = (2k - 1)(n-k) $. Let $ n = 2m $ $$ \begin{align*}\prod_{k = 1}^{\frac{n}{2}} (2k - 1)(n- k) &= \prod_{k = 1}^{m} (2k - 1) \prod_{k=1}^m (2m - k) \\ &=(2m-1)!! \frac{(2m-1)!}{(m-1)!} \end{align*}$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.