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Math Help - Series and Sequences help

  1. #1
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    Series and Sequences help

    My book already tells me the answer to a problem and does it step by step, but I have a question b/w a step.

    The original question is here:

    Sigma from 1 to infinity of 3^(2k)*5^(1-k)

    After that, the book says to simplify it to this:

    (9^k)/(5^k-1)

    After that, this is where I have my problem. It say to simplify it once more to what I have beloew, only this does not make sense to me. How do you break up a 9^k into a 9 by itself? Furthermore, why is a 9 put into the 1/5? Wouldn't it just be 1/5^k-1 times 9^k? I really do not understand simplifying these types of equations so if anyone could help me with this one or link me to a site that shows how to simplify these things, I would greatly appreciate it. The only rule I know about any number ^k is that anything ^k+1 is the same thing as anything times anything ^k. Is there a list of these rules somewhere dealing with numbers to the power of a variable?

    9*(9/5)^k-1
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  2. #2
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    {9^k \over 5^{k-1}} = {9^{1+k-1} \over 5^{k-1}} = {9^1\cdot 9^{k-1}\over 5^{k-1}} = 9 \cdot(\frac95)^{k-1}

    edit: x^{a+b} = x^a x^b and x^0=1 are the defining property of exponents. Letting a=-b you get 1 = x^0=x^{a+(-a)} = x^a x^{-a} so x^{-a} = 1/x^a. Also, x^1 = x^{\frac1n+\cdots+\frac1n} = (x^{1/n})\cdots(x^{1/n}) = (x^{1/n})^n so x^{1/n} = \sqrt[n]{x^1}. You can look the others up on wikipedia I think.
    Last edited by maddas; April 9th 2010 at 05:08 PM.
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  3. #3
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    Quote Originally Posted by maddas View Post
    {9^k \over 5^{k-1}} = {9^{1+k-1} \over 5^{k-1}} = {9^1\cdot 9^{k-1}\over 5^{k-1}} = 9 \cdot(\frac95)^{k-1}
    Wow, I never would have gotten that. That's too far outside the box for me. Thank you so much.
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  4. #4
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    Quote Originally Posted by Simon777 View Post
    My book already tells me the answer to a problem and does it step by step, but I have a question b/w a step.

    The original question is here:

    Sigma from 1 to infinity of 3^(2k)*5^(1-k)

    After that, the book says to simplify it to this:

    (9^k)/(5^k-1) \textcolor{red}{= \frac{5}{5} \cdot \frac{9^k}{5^{k-1}} = 5 \cdot \left(\frac{9}{5}\right)^k}

    \textcolor{red}{5 \sum_{k=1}^\infty \left(\frac{9}{5}\right)^k}

    ... which diverges.
    ...
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