[UNRESOLVED] --->Combinations/Sum

I'd like to see some other solutions to this question. Evaluate the following:
http://vbforums.com/attachment.php?s=&postid=1480938

noone?

SilverSprite, you posted:

I'd like to see some other solutions to this question. Evaluate the following:
(Your function here)

Attachment: math.gif
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There havebeen over 10 viewers of your posting, and none have answered. I, for one, would like to see the solution(s) thatyou imply with "I'd like to see some other solutions to this question". Maybe, then, others will take more interest and try to comply.

Ok here is the solution I know of in two parts. The first part is that of Andrei Jorza:

(2^k)(2n-k C k) is the coefficient of x^n in the expansion of (2^k)(1+x)^(2n-k). If we add the coefficients we have ((1+x)^(2n+1) - (2^(n+1))(1+x)^n)/(x-1) = ((2^(n+1))(1+x)^n - (1+x)^(2n+1))(1 + x + x^2 + ...). Here the coefficient of x^n is 2^(n+1) * sum from k=0 to n of nCk - sum from k=0 to n of (2n+1)Ck = 2^(2n+1) - 2^(2n) = 2^(2n).

(note: nCk reads n choose k)

Now the second part I understand even less.. I don't know why it works but notice that from above we have (2^k)*(2n-k C k) is the coefficient of x^n in the expansion of (2^k)*(1+x)^(2n-k). If we sum the coefficients by substituting x=1 we have 2^k * 2^(2n-k) = 2^(2n).