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clubcanthandlekg

  • 2 years ago

let n be a positive integer. prove that: 1(nC1) + 2(nC2) +...+n(nCn) = n2^(n-1)

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  1. cnknd
    • 2 years ago
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    i'll rewrite the original expression in 2 ways (and i'll call it X): X = 0(nC0) + 1(nC1) + ... + (n-1)(nCn-1) + n(nCn) X = n(nCn) + (n-1)(nCn-1) + .... + 1(nC1) + 0(nC0) now remember nCn = nC0, nC1 = nCn-1, ... so if I add up the 2 equations I wrote, what do I get?

  2. jishan
    • 2 years ago
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    |dw:1353851927535:dw|

  3. jishan
    • 2 years ago
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    |dw:1353852046696:dw|

  4. jishan
    • 2 years ago
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    |dw:1353852202839:dw|

  5. FoolIsHere
    • 2 years ago
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    Another possible approach, will be to take the binomial expansion of \( (1+x)^n \). Differentiate both sides with respect to \( x \), then substitute \( x=1 \).

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