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Math Help - fractions with powers

  1. #1
    Member Jskid's Avatar
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    fractions with powers

    How does 1-\frac{1}{3^{k-1}}+\frac{2}{3^k}=1- \frac{1}{3^k}?
    I tried putting it over a common denominator and got 1-\frac{3^k+6^{k-1}}{3^{k-1} \cdot 3^k}
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  2. #2
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    Re: fractions with powers

    Quote Originally Posted by Jskid View Post
    How does 1-\frac{1}{3^{k-1}}+\frac{2}{3^k}=1- \frac{1}{3^k}?
    I tried putting it over a common denominator and got 1-\frac{3^k+6^{k-1}}{3^{k-1} \cdot 3^k}
    1-\frac{1}{3^{k-1}}+\frac{2}{3^k}=1-\frac{1}{3^{k-1}}+\frac{2}{3\cdot 3^{k-1}}=1-\left(\frac{1}{3^{k-1}}-\frac{2}{3\cdot 3^{k-1}}\right)=

    =1-\left(\frac{3-2}{3\cdot 3^{k-1}}\right)=1-\frac{1}{3^k}
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  3. #3
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    Re: fractions with powers

    Hello, Jskid!

    A slightly different approach . . .


    \text{How does }\,1-\frac{1}{3^{k-1}}+\frac{2}{3^k}\:=\:1- \frac{1}{3^k}\:?

    Consider the two fractions: . - \frac{1}{3^{k-1}} + \frac{2}{3^k}

    Multiply the first fraction by \tfrac{3}{3}\!:\;\;-\frac{1}{3^{k-1}}\cdot\frac{3}{3}\;+\;\frac{2}{3^k}

    Then we have: . -\frac{3}{3^k} + \frac{2}{3^k} \;=\;\frac{-3 + 2}{3^k} \;=\;-\frac{1}{3^k}

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