Re: Maths problem
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Has anyone thought of a combinatorial solution to this ?
I have tried and failed .
We wish to prove :
2nCn + 2c1*(2n-2)C(n-1) + 4c2*(2n-4)C(n-2) + ...+ 2nCn = 4^n
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I don't know of one. I was thinking of the generating function method. But it would be interesting to know.
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