## Statistics summation question:

This equation comes out of deriving the canonical partition function for some system. However, the question is more math based. I am having trouble understanding the simplification that was performed in the text:

∑ from N=0 to M of: (M!exp((M-2N)a))/(N!(M-N)!) supposedly becomes

exp(Ma)(1+exp(-2a))^M.... I tried to look at the first few terms and see how I can simplify this, and no dice... Anyone have any ideas? a is just a constant.

Also, the next step is that the above becomes (2cosh(a))^M Which is great, except, huh? I'm more of a scientist than a math person, so I apologize if I am missing something elementary.

Thanks!
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 Recognitions: Homework Help Remember that $$\sum_{n=0}^m \frac{m!}{n!(m-n)!} 1^{(m-n)} b^n = (1 + b)^m$$ In your sum $$\exp\left((M-2N)a\right) = \exp\left(Ma\right) \cdot \left(\exp(-2a)\right)^N$$ so your sum is $$\sum_{N=0}^M {\frac{M!}{N!(M-N)!} \exp\left((M-2N)a\right)} = e^{Ma} \sum_{N=0}^M {\frac{M!}{N!(M-N)!} \left(e^{-2a}\right)^N = e^{Ma} \left(1 + e^{-2a}\right)^M$$ For the second one - look at the definition of the hyperbolic function as exponentials, and rearrange terms.
 Thanks so much!! That really clears things up.

 Tags factorials, partition function, simplification, summation