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anonymous
 4 years ago
2n C 0 + 2n C 1 + 2n C 2 +....................2n C n = 2^(2n1)..is it true.
if yes or no, how??
anonymous
 4 years ago
2n C 0 + 2n C 1 + 2n C 2 +....................2n C n = 2^(2n1)..is it true. if yes or no, how??

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What is C? Some constant? It's missing from the RHS..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0n C r = {n!/(nr)!r!} now, can u understand my question?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i am not getting you...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's the same thing, n choose k is a combination e.g. binomial coefficient

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it's just another way of saying it :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It is false. It should be 2^n I think.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Take n=1 then 4 c 0 + 4 c 1 = 1 + 4 = 5 but 2^(21) is not 5.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No it should not be 2^n

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You are correct it is not 2^n. I was thinking sum k=0 to n of n c k but that expression is sum k=0 to n of 2n c k

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I was speculating what the correct formula was...I said 2^n since sum k=0 to n of n c k = 2^n is a common identity but I forgot your expression was 2n c k.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Just ignore that post :) The identity is false. Just take n=1 for a counter example.
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