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Thread: Series proof.

  1. #1
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    Sum proof.

    Hello, I am stuck with the following problem.

    Verify the following identity for $\displaystyle n \geq 2$.

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

    I tried an induction proof, but I couldn't get it right. Do I need to consider when n is odd and when n is even?

    Thanks in advance
    Last edited by akolman; Sep 13th 2009 at 07:31 PM. Reason: wrong title =(
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  2. #2
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    this is a sum, not a series, and the solution is quite easy:

    $\displaystyle \sum\limits_{k=0}^{n}{\binom nk(-1)^{k}1^{n-k}}=(1-1)^{n}=0^{n}=0.$
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  3. #3
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    Quote Originally Posted by akolman View Post
    Hello, I am stuck with the following problem.

    Verify the following identity for $\displaystyle n \geq 2$.

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

    I tried an induction proof, but I couldn't get it right. Do I need to consider when n is odd and when n is even?

    Thanks in advance
    Like Krizalid said, this is an example of the Binomial Theorem.

    If you know that

    $\displaystyle (x + y)^n = \sum_{k = 0}^n {\left(_k^n\right) x^{n - k} y^k}$

    it is relatively easy to see that if you let $\displaystyle x = 1$ and $\displaystyle y = -1$ you get

    $\displaystyle (1 - 1)^n = \sum_{k = 0}^n {\left(_r^n\right) (1)^{n - k} (-1)^k}$

    $\displaystyle 0^n = \sum_{k = 0}^n {\left(_k^n\right) (-1)^k}$

    Therefore $\displaystyle \sum_{k = 0}^n {\left(_k^n\right) (-1)^k} = 0$.
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