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Math Help - Mathematical Induction

  1. #1
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    Mathematical Induction

    Hi I hope someone can help me with this problem:

    Prove the following formula by induction:

    1+(1/2)+(1/4)+(1/8)+...+1/(2^n) = 2 - 1/(2^n)

    Thanks in advance
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by storchfire1X View Post
    Hi I hope someone can help me with this problem:

    Prove the following formula by induction:

    1+(1/2)+(1/4)+(1/8)+...+1/(2^n) = 2 - 1/(2^n)

    Thanks in advance
    A) Base case n=0, then the left hand side is 1, and the right hnd side is 1.

    B) Suppose this true for some n=k \in \mathbb{N_+}

    Then:

    <br />
[1+1/2+1/4+1/8+...+1/2^{k}]+1/2^{k+1} = [2 - 1/2^k]+1/2^{k+1} = 2-1/2^{k+1}<br />

    hence it is true for k+1.

    Thus combining A), and B) we have proven that it is true for all n \in \mathbb{N} by mathematical induction

    RonL
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  3. #3
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    Quote Originally Posted by storchfire1X View Post
    Hi I hope someone can help me with this problem:

    Prove the following formula by induction:

    1+(1/2)+(1/4)+(1/8)+...+1/(2^n) = 2 - 1/(2^n)

    Thanks in advance
    n = 0

    1 = 2 - \frac{1}{2^0} = 2 - 1 = 1 (Check!)

    Now assume this is true for some n = k. We wish to show that it is true for n = k + 1.

    So assume:
    1 + \frac{1}{2} + ~ ... ~ + \frac{1}{2^k} = 2 - \frac{1}{2^k}
    is true.

    1 + \frac{1}{2} + ~ ... ~ + \frac{1}{2^k} + \frac{1}{2^{k + 1}}

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

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

    = 2 - \left ( \frac{1}{2^k} - \frac{1}{2} \cdot \frac{1}{2^k} \right )

    = 2 - \left ( 1 - \frac{1}{2} \right ) \frac{1}{2^k}

    = 2 - \frac{1}{2} \cdot \frac{1}{2^k}

    = 2 - \frac{1}{2^{k + 1}}
    as we needed to show.

    -Dan
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