Proving the Evenness of (\stackrel{2n}{n}) Using the Binomial Theorem

ipitydatfu
Messages
13
Reaction score
0

Homework Statement



prove that (\stackrel{2n}{n}) is even when n \geq1

Homework Equations



as a hint they gave me this identity:
\stackrel{n}{k}= (n/k)(\stackrel{n-1}{k-1})

The Attempt at a Solution



by using that identity i got:

(\stackrel{2n}{n}) = (2n/n) (\stackrel{2n-1}{n-1})
= (2) (\stackrel{2n-1}{n-1})

i thought anything multiplied by 2 is an even number. but then again this is discrete math. how would i inductively show that this is true?
 
Physics news on Phys.org
That's pretty much it. The definition of "even number" is that it is of the form 2k for some integer k. Do you already know that \left(\begin{array}{c}n\\2i\end{array}\right) is always an integer?
 
oh yeah! I forgot about that! thanks!
 
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