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Math Help - using mathematical induction to prove..

  1. #1
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    using mathematical induction to prove..

    hi guys having trouble proving this result.

     <br />
\sum_{k=0}^n(-\frac{1}{2})^k =\frac{2}{3} + \frac{1}{3}\frac{(-1)^n}{2^n},

    for all n \in \mathbb{N}

    Any help much appreciated.. thanks
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  2. #2
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    Quote Originally Posted by jvignacio View Post
    hi guys having trouble proving this result.

     <br />
\sum_{k=0}^n(-\frac{1}{2})^k =\frac{2}{3} + \frac{1}{3}\frac{(-1)^n}{2^n},

    for all n \in \mathbb{N}

    Any help much appreciated.. thanks
    First show the base case is true n=1

     <br />
\sum_{k=0}^{1}(-\frac{1}{2})^k =1-\frac{1}{2}=\frac{1}{2}?=\frac{2}{3} + \frac{1}{3}\frac{(-1)^1}{2^1}=\frac{1}{2},

    so it checks. Now assume it is true for n

     <br />
\sum_{k=0}^n(-\frac{1}{2})^k =\frac{2}{3} + \frac{1}{3}\frac{(-1)^n}{2^n},

    no show N+1

     <br />
\sum_{k=0}^{n+1}(-\frac{1}{2})^k=\left(-\frac{1}{2}\right)^{n+1}+\sum_{k=0}^{n}(-\frac{1}{2})^k

    Now by the induction hypothesis we get

    \left(-\frac{1}{2}\right)^{n+1} +\frac{2}{3} + \frac{1}{3}\frac{(-1)^n}{2^n}

    \frac{2}{3}+\left(-\frac{1}{2}\right)^{n+1}+\frac{1}{3}\frac{(-1)^n}{2^n}=\frac{2}{3}+\left(-\frac{1}{2}\right)\frac{(-1)^{n}}{2^n}+\frac{1}{3}\frac{(-1)^n}{2^n}

    \frac{2}{3}+\left[\left( -\frac{1}{2}\right) +\frac{1}{3}\right]\frac{(-1)^n}{2^n}=\frac{2}{3}-\left(\frac{1}{6} \right)\frac{(-1)^{n}}{2^{n}}

    \frac{2}{3}-\left(\frac{1}{6} \right)\frac{(-1)^{n}}{2^{n}}\left( \frac{-1}{2}\right)\left( \frac{2}{-1}\right)=\frac{2}{3}+\left( \frac{2}{3}\right) \frac{(-1)^{n+1}}{2^{n+1}}

    This shows n+1 so we are done
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    First show the base case is true n=1

     <br />
\sum_{k=0}^{1}(-\frac{1}{2})^k =1-\frac{1}{2}=\frac{1}{2}?=\frac{2}{3} + \frac{1}{3}\frac{(-1)^1}{2^1}=\frac{1}{2},

    so it checks. Now assume it is true for n

     <br />
\sum_{k=0}^n(-\frac{1}{2})^k =\frac{2}{3} + \frac{1}{3}\frac{(-1)^n}{2^n},

    no show N+1

     <br />
\sum_{k=0}^{n+1}(-\frac{1}{2})^k=\left(-\frac{1}{2}\right)^{n+1}+\sum_{k=0}^{n}(-\frac{1}{2})^k

    Now by the induction hypothesis we get

    \left(-\frac{1}{2}\right)^{n+1} +\frac{2}{3} + \frac{1}{3}\frac{(-1)^n}{2^n}

    \frac{2}{3}+\left(-\frac{1}{2}\right)^{n+1}+\frac{1}{3}\frac{(-1)^n}{2^n}=\frac{2}{3}+\left(-\frac{1}{2}\right)\frac{(-1)^{n}}{2^n}+\frac{1}{3}\frac{(-1)^n}{2^n}

    \frac{2}{3}+\left[\left( -\frac{1}{2}\right) +\frac{1}{3}\right]\frac{(-1)^n}{2^n}=\frac{2}{3}-\left(\frac{1}{6} \right)\frac{(-1)^{n}}{2^{n}}

    \frac{2}{3}-\left(\frac{1}{6} \right)\frac{(-1)^{n}}{2^{n}}\left( \frac{-1}{2}\right)\left( \frac{2}{-1}\right)=\frac{2}{3}+\left( \frac{2}{3}\right) \frac{(-1)^{n+1}}{2^{n+1}}

    This shows n+1 so we are done
    very well explained. much appreciated
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    First show the base case is true n=1

    \sum_{k=0}^{1}(-\frac{1}{2})^k =1-\frac{1}{2}

    quick question how did u get 1-\frac{1}{2} the 1 - ?
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  5. #5
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    Quote Originally Posted by jvignacio View Post
    \sum_{k=0}^{1}(-\frac{1}{2})^k =1-\frac{1}{2}

    quick question how did u get 1-\frac{1}{2} the 1 - ?
    plug in k=0 and k=1 into the sum, see what you get.
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