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Math Help - Sequences and Series

  1. #1
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    Sequences and Series

    Q.: Please see attachment.

    In this case, the text book has included the answer. I have 3^n/2, they have just 3^n. Can someone help me figure out how to factor out the 2? Thank you.
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  2. #2
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    Re: Sequences and Series

    Quote Originally Posted by GrigOrig99 View Post
    Q.: Please see attachment.

    In this case, the text book has included the answer. I have 3^n/2, they have just 3^n. Can someone help me figure out how to factor out the 2? Thank you.
    You're very close, it is that last simplification that is wrong - the laws of exponents do not work like that. Use the law that says a^{b+c} = a^ba^c to express 3^{n+1} in terms of 3^n



    Spoiler:
    \dfrac{3^{n+1} - 3^n}{2} = \dfrac{3(3^n) - 3^n}{2} = \dfrac{3^n(3-1)}{2} = 3^n
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  3. #3
    Super Member TheChaz's Avatar
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    Re: Sequences and Series

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


    edit: the biggest spoiler was that someone posted before me!

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  4. #4
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    Re: Sequences and Series

    Thanks for the help guys. I really appreciate it.
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