Question regarding Binomial Theorem

mess1n
Messages
24
Reaction score
0
Hey people, I've racked my brain on this question for hours and can't seem to get to grips with it, wondering if i could get a little guidance?

Homework Statement



Considering the co-efficient of x^n in the expansion of (1+x)^n(1+x)^n, show:

[the sum from k=0 -> k=n of:] [nCk]2 = 2nCn

The Attempt at a Solution



i've found the co-efficient of x^n by using the binomial expansion on (1+x)^2n, and i can see that the binomial expansion of (1+x)^2n will give a series of squared co-efficients, but I have no idea how to relate the two :s.

Any help would be much appreciated.

Cheers,
Andrew
 
Physics news on Phys.org
Can't you use induction?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top