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Re: Maths problem
[ QUOTE ]
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 [/ QUOTE ] 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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