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Suppose n is even, prove:

[tex]\sum[/tex]k=0->n/2, C(n,2k)=2^(n-1)=[tex]\sum[/tex]k=1->n/2, C(n,2k-1)

Give a combinatorial argument to prove that: (I've figured out this one...)

[tex]\sum[/tex]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 [tex]\sum[/tex]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!

[tex]\sum[/tex]k=0->n/2, C(n,2k)=2^(n-1)=[tex]\sum[/tex]k=1->n/2, C(n,2k-1)

Give a combinatorial argument to prove that: (I've figured out this one...)

[tex]\sum[/tex]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 [tex]\sum[/tex]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!

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