How to proof; a sumformula

1. Nov 14, 2009

Vince00

1. The problem statement, all variables and given/known data
Proof that $$\sum a, m=0,$$ (a-m)! / m!(a-2m)! = the fibonacci sequence.
2. Relevant equations
Fibonacci: 1, 1, 2, 3, 5, ... (but I think everyone knows that one!)
3. The attempt at a solution
Let Xm = $$\suma m=0$$ (a-m)! / m!(a-2m)!
I think I better proof Xm+2 = Xm+1 + Xm (follows from the fibonacci), so I can conclude that Xm = Fm+1.
I think using induction is the best way to go: Prooving it for X0 and X1, maybe even X2.
My attempt:
$$\sum a+2, m=0,$$ (a+2-m)! / m!(a+2-2m)! = $$\sum a+1, m=0,$$ (a+1-m)! / m!(a+1-2m)! + $$\sum a, m=0,$$ (a-m)! / m!(a-2m)!
=$$\sum a+1, m=0,$$ (a+1-m)! / m!(a+1-2m)! + $$\sum a+1, m=1,$$ (a-m-1)! / (m-1)!(a-m-1-m+1)! ... but I think I am making big mistakes here, because whatever I do, I get stuck.
I just don't see it!

Vince, freshmen physics, sorry but I don't know how to use the symbols and stuff!

2. Nov 15, 2009

n!kofeyn

It's difficult to read what you have written. Try editing it by using the proper LaTeX. Just click on my output below to see the way to do it.
$$\sum_{n=0}^\infty a_n$$ for sums and subscripts
$$\frac{(a-m)!}{m!(a-2m)!}$$ for fractions
Also, what is Xm?

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook