Proving Binomial Coefficients for Even n | Combinatorial Argument

[jbear12]

Suppose n is even, prove:
$$\sum$$k=0->n/2, C(n,2k)=2^(n-1)=$$\sum$$k=1->n/2, C(n,2k-1)
Give a combinatorial argument to prove that: (I've figured out this one...)
$$\sum$$k=1->n, C(n,k)^2=C(2n,n)

For the first problem, I tried to break C(n, 2k) into C(n+1,2k)-C(n, 2k-1), but they didnt seem to work very well. (I also noticed that $$\sum$$0, n C(n,2k)=2^n, but I can't really use this to solve the problem.)
For the second problem, I don't have a clue.
Any thoughts?
Thanks!

 

[HallsofIvy]

You want to prove that, for n even, $\sum_{k= 0}^{n/2}C(n, 2k)= 2^{n-1}$ and that $\sum_{k=0}^{n/2} C(n, 2k-1)= 2^{n-1}$

You think mean you have tried to use C(n, 2k)= C(n-1,2k)-C(n-1, 2k-1) . That works nicely! If you take i= 2k, That just says C(n, i)= C(n-1, i)+ C(n-1, i) and your sum is $\sum_{k= 0}^{n/2}C(n, 2k)= \sum_{i= 0}^n C(n-1, i)= 2^{n-1}$!

For the second one, let i= 2k-1.

 

[jbear12]

You want to prove that, for n even, $\sum_{k= 0}^{n/2}C(n, 2k)= 2^{n-1}$ and that $\sum_{k=0}^{n/2} C(n, 2k-1)= 2^{n-1}$

You think mean you have tried to use C(n, 2k)= C(n-1,2k)-C(n-1, 2k-1) . That works nicely! If you take i= 2k, That just says C(n, i)= C(n-1, i)+ C(n-1, i) and your sum is $\sum_{k= 0}^{n/2}C(n, 2k)= \sum_{i= 0}^n C(n-1, i)= 2^{n-1}$!

For the second one, let i= 2k-1.

Thanks HallsofIvy.
But I don't get it how you get from C(n, 2k)= C(n-1,2k)-C(n-1, 2k-1) to C(n, i)= C(n-1, i)+ C(n-1, i). Taking i=2k, why is C(n-1,i-1) equal to -C(n-1,i).
Also, why is $\sum_{i= 0}^n C(n-1, i)= 2* \sum_{i= 0}^{n/2} C(n-1, i)$