anonymous
  • anonymous
2n C 0 + 2n C 1 + 2n C 2 +....................2n C n = 2^(2n-1)..is it true. if yes or no, how??
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
What is C? Some constant? It's missing from the RHS..
anonymous
  • anonymous
combination sign..
anonymous
  • anonymous
n choose k?

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anonymous
  • anonymous
n C r = {n!/(n-r)!r!} now, can u understand my question?
anonymous
  • anonymous
yes, it's n choose k
anonymous
  • anonymous
i am not getting you...
anonymous
  • anonymous
It's the same thing, n choose k is a combination e.g. binomial coefficient
anonymous
  • anonymous
it's just another way of saying it :)
anonymous
  • anonymous
okay..
anonymous
  • anonymous
No.
anonymous
  • anonymous
It is false. It should be 2^n I think.
anonymous
  • anonymous
Take n=1 then 4 c 0 + 4 c 1 = 1 + 4 = 5 but 2^(2-1) is not 5.
anonymous
  • anonymous
No it should not be 2^n
anonymous
  • anonymous
You 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
  • anonymous
2^(2n-1).
anonymous
  • anonymous
I 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
  • anonymous
Just ignore that post :) The identity is false. Just take n=1 for a counter example.

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