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Math Help - Proving Binomial Theorem (1-1)^n = 0

  1. #1
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    Proving Binomial Theorem (1-1)^n = 0

    This was the best fit for Binomial Theorem I could find--Probability & Stats, hopefully it's in the right spot.

    The question:
    Use the binomial expansion of (a + b)^n and the identity (1 - 1)^n = 0 to prove
    (n choose 0) - (n choose 1) + (n choose 2) - ... + (-1)^2(n choose n) = 0, where n != 0

    I perfectly understand the varying signs and why it works out like that in my head but I can't prove it. My proving sucks bad. I think in the obvious common sense way that's usually wrong because I assume a buncha things.

    I'm thinking:
    (n choose 0) - (n choose 1) + (n choose 2) - ... + (-1)^2(n choose n) = 0
    R.S.
    Multiply it by a.
    a(n choose 0) - a(n choose 1) + a(n choose 2) - ... + a(-1)^2(n choose n)
    Through inspection
    (a - a)^n
    (0)^n
    0
    L.S. = R.S.
    0 = 0

    But I'm pretty sure that's wrong... Any ideas/suggestions =)
    And expect more from me in the near future with random pesterings of more proving questions cause I'm super hopeless at those.

    Thanks a lot in advance.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by blaise View Post
    This was the best fit for Binomial Theorem I could find--Probability & Stats, hopefully it's in the right spot.

    The question:
    Use the binomial expansion of (a + b)^n and the identity (1 - 1)^n = 0 to prove
    (n choose 0) - (n choose 1) + (n choose 2) - ... + (-1)^2(n choose n) = 0, where n != 0

    I perfectly understand the varying signs and why it works out like that in my head but I can't prove it. My proving sucks bad. I think in the obvious common sense way that's usually wrong because I assume a buncha things.

    I'm thinking:
    (n choose 0) - (n choose 1) + (n choose 2) - ... + (-1)^2(n choose n) = 0
    R.S.
    Multiply it by a.
    a(n choose 0) - a(n choose 1) + a(n choose 2) - ... + a(-1)^2(n choose n)
    Through inspection
    (a - a)^n
    (0)^n
    0
    L.S. = R.S.
    0 = 0

    But I'm pretty sure that's wrong... Any ideas/suggestions =)
    And expect more from me in the near future with random pesterings of more proving questions cause I'm super hopeless at those.

    Thanks a lot in advance.
    By the binomial theorem:

    (1-x)^n=\sum_{k=0}^n {n \choose k}(-x)^k

    So when x=1, we have:

    (1-1)^n=\sum_{k=0}^n {n \choose k}(-1)^k=0

    which is the required result.

    RonL
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