# Binomial Theorem Problem

• Aug 27th 2008, 04:26 AM
Number Cruncher 20
Binomial Theorem Problem
Hello,

I am having trouble deriving a simple closed form expression for the following sum:

Sn =

http://www.mathhelpforum.com/math-he...488598ba-1.gif for n = 0,1,2,3,4,.....

I am not sure how to use the binomial theorem to derive the expression. Any help would be much appreciated.
• Aug 27th 2008, 04:30 AM
flyingsquirrel
Hello,

Your sum can be written $\displaystyle S_n=\sum_{i=0}^n\binom{n}{i}(-1)^i1^{n-i}$ which looks like $\displaystyle \sum_{i=0}^n\binom{n}{i}x^iy^{n-i}=(x+y)^n$, doesn't it ?
• Aug 27th 2008, 04:48 AM
Number Cruncher 20
Binomial Theorem Problem
Yes I see it now, thanks. By any chance would you know how to get started on proving that:

http://www.mathhelpforum.com/math-he...fb3cf705-1.gif for http://www.mathhelpforum.com/math-he...195e4ccc-1.gif.
• Aug 27th 2008, 05:01 AM
flyingsquirrel
Quote:

Originally Posted by Number Cruncher 20
By any chance would you know how to get started on proving that:

http://www.mathhelpforum.com/math-he...fb3cf705-1.gif for http://www.mathhelpforum.com/math-he...195e4ccc-1.gif.

This is straightforward. We have

$\displaystyle \binom{n}{k}\binom{k}{l}=\frac{n!}{k!(n-k)!}\cdot \frac{k!}{l!(k-l)!}$ and $\displaystyle \binom{n}{l}\binom{n-l}{k-l}=\frac{n!}{l!(n-l)!}\cdot \frac{(n-l)!}{(k-l)!(n-l-(k-l))!}$

If you simplify these two expressions you'll see that they are equal.