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Math Help - Mathematical Induction

  1. #1
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    Mathematical Induction

    Prove by mathematical induction

    1 + 3 + 3^2 + ... + 3^(n-1) = [3^(n-1)] / 2

    .
    i really need helpp, i really don't get it after the second step.
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  2. #2
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    Quote Originally Posted by phillychum View Post
    Prove by mathematical induction

    1 + 3 + 3^2 + ... + 3^(n-1) = [3^(n-1)] / 2

    .
    i really need helpp, i really don't get it after the second step.
    Hi phillychum,

    it should read 1+3+3^2+....+3^{n-1}=\frac{3^n-1}{2}

    If this is true, then

    1+3+3^2+......+3^{n-1}+3^n should =\frac{3^{n+1}-1}{2}

    We can prove if this is the case by adding 3^n to both sides of the original equality.

    1+3+3^2+....+3^{n-1}+3^n=\frac{3^n-1}{2}+3^n\ ?

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

    This means that if 1+3+3^2+...+3^{n-1}=\frac{3^n-1}{2}

    then 1+3+3^2+...+3^n=\frac{3^{n+1}-1}{2}

    Hence the inductive step is proven.

    Now it only needs to be proven for term number 1.
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