Even vs. Odd in a combinatoric proof

I was given a problem to simplify the following:

$\displaystyle \sum_{k \geq 0} \binom{n}{2k} $

where:

$\displaystyle \binom{n}{j} = 0 $ whenever j > n.

It's easy to see that the problem is equivalent to:

$\displaystyle \sum_{k = 0}^n \binom{n}{2k} $

Now I can see that for odd values of n, this expands to:

$\displaystyle \binom{n}{0} + \binom{n}{2} + ... + \binom{n}{n-1} $

and since:

$\displaystyle \binom{n}{n-1} = \binom{n}{1}, \binom{n}{n-3} = \binom{n}{3}, ... $

we can use the last n/2 terms to "fill in the gaps", simplifying it to be:

$\displaystyle \frac{\sum_{k=0}^n \binom{n}{k}}{2} = \frac{2^n}{2} = 2^n-1 $

That's easy. But I don't know why the proof still works for even numbers of n. You no longer have the same correspondence (n choose 0 no longer has the equivalent n choose n-1 as a term):

$\displaystyle \binom{n}{0} + \binom{n}{2} + ... + \binom{n}{n-2} + \binom{n}{n} $

So it can't "fill in the gaps" in the same way.

Can anyone figure this out? I've been at this for hours and the explanations I'm being given by the texts are just crap.

correction to the previous post

forget what i've mentioned in my previous post it's all wrong and really sorry abount it.(because n is a given value it can either be odd or even)(Headbang)

but check out this method. Hope it's not wrong too.

$\displaystyle (a+b)^n=\sum_{r=0}^n\binom{n}{r}a^rb^{n-r}$

if n is odd

if a=1,b=-1

$\displaystyle 0=\sum_{r=0}^n\binom{n}{2r}-\sum_{r=0}^n\binom{n}{2r+1}$ ___ (1)

if a=b=1;

$\displaystyle 2^n=\sum_{r=0}^n\binom{n}{2r}-\sum_{r=0}^n\binom{n}{2r+1}$___(2)

by (1)+(2) i think we'll get the result for odd n

when n is even, same 2 equations could be achieved.