By expanding (x/(1 - x))^n , 0<x<1 , in 2 ways, prove that: nC0(2n-1)Cn - nC1(2n - 2)Cn + nC2(2n - 3)Cn - ... = 1 NOTE: nC0 represents n choose 0. nC0(2n-1)Cn represents (n choose 0) multiplied by (2n - 1 choose n)
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Originally Posted by fardeen_gen By expanding (x/(1 - x))^n , 0<x<1 , in 2 ways, prove that: nC0(2n-1)Cn - nC1(2n - 2)Cn + nC2(2n - 3)Cn - ... = 1 NOTE: nC0 represents n choose 0. nC0(2n-1)Cn represents (n choose 0) multiplied by (2n - 1 choose n) Use the binomial expansion . Then . But also , and so (using the above expansion for ). Now compare the coefficients of in the two expansions of .
Last edited by Opalg; February 10th 2009 at 02:11 AM. Reason: typo
That gave me: 1 = nC0.(n-1)Cn - nC1.nCn + nC2.(n + 1)Cn + ... Which is obviously wrong. What am I missing?
Originally Posted by fardeen_gen That gave me: 1 = nC0.(n-1)Cn - nC1.nCn + nC2.(n + 1)Cn + ... Which is obviously wrong. What am I missing? It's not wrong. It's a finite series, which looks like . Now write the terms in reverse order, and use the fact that .
What is (n - 1)Cn?
Originally Posted by fardeen_gen What is (n - 1)Cn? I think you have to interpret that as 0. You can check that by tracing this term back to where it came from in the expansion of .
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