Trenters4325
Junior Member
- Joined
- Apr 8, 2006
- Messages
- 122
To prove Vandermonde's identity
with (1+x)^m*(1+x)^n = (1+x)^(m+n),
you will get something like
\(\displaystyle \sum^{m}_{a=0}\sum^{n}_{b=0}C(m,a)C(m,b)*x^{a+b} = \sum^{m+n}_{c=0}C(m+n,c)x^c\).
I see how you could set a+b = c, but I am having trouble rigorously justifying how you get rid of the summations?
with (1+x)^m*(1+x)^n = (1+x)^(m+n),
you will get something like
\(\displaystyle \sum^{m}_{a=0}\sum^{n}_{b=0}C(m,a)C(m,b)*x^{a+b} = \sum^{m+n}_{c=0}C(m+n,c)x^c\).
I see how you could set a+b = c, but I am having trouble rigorously justifying how you get rid of the summations?