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clubcanthandlekg
Group Title
let n be a positive integer. prove that:
1(nC1) + 2(nC2) +...+n(nCn) = n2^(n1)
 2 years ago
 2 years ago
clubcanthandlekg Group Title
let n be a positive integer. prove that: 1(nC1) + 2(nC2) +...+n(nCn) = n2^(n1)
 2 years ago
 2 years ago

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cnknd Group TitleBest ResponseYou've already chosen the best response.0
i'll rewrite the original expression in 2 ways (and i'll call it X): X = 0(nC0) + 1(nC1) + ... + (n1)(nCn1) + n(nCn) X = n(nCn) + (n1)(nCn1) + .... + 1(nC1) + 0(nC0) now remember nCn = nC0, nC1 = nCn1, ... so if I add up the 2 equations I wrote, what do I get?
 2 years ago

jishan Group TitleBest ResponseYou've already chosen the best response.0
dw:1353851927535:dw
 one year ago

jishan Group TitleBest ResponseYou've already chosen the best response.0
dw:1353852046696:dw
 one year ago

jishan Group TitleBest ResponseYou've already chosen the best response.0
dw:1353852202839:dw
 one year ago

FoolIsHere Group TitleBest ResponseYou've already chosen the best response.0
Another possible approach, will be to take the binomial expansion of \( (1+x)^n \). Differentiate both sides with respect to \( x \), then substitute \( x=1 \).
 one year ago
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