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Thread: have difficulties with Geometric series questions

  1. #1
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    have difficulties with Geometric series questions

    Question:
    have difficulties with Geometric series questions-qwq.jpg

    I attempted to answer it.

    have difficulties with Geometric series questions-gh1.jpg

    have difficulties with Geometric series questions-gh2.jpg

    The problem I have is with the worked solutions is how did the author come up with:
     \frac{1}{a}(\frac{r^{n-1}+r^{n-1}+r^{n-2}+r^{n-31}+...+r^{0}}{r^{(n-1)(n-2)(n-3)..3.2.1}})

    I thought that  \frac{1}{a}+\frac{1}{ar} + ....+ \frac{1}{ar^{n-1}} = \frac{1}{a}(r^{0}+r^{-1}+r^{-2}+ r^{-3})

    therefore r = common ratio = r^{-1}


    and lastly how did the author need up with  \frac{a^{n-1}n(n-1)}{2P_{n}} where  P_{n}  = a^{n}\frac{r^{n(n-1)/2}}{2} as the answer


    I thought  P_{n}  = a^{n}\frac{r^{n(n-1)/2}}{2} is wrong.
    Shouldn't  P_{n} = a^{n}r^{\frac{n(n-1)}{2}} ?

    Please if someone can be kind to explain to me thank you


    Edit: Please don't worry about the second part of the question
     \frac{1}{(a)(ar).....(ar^{n-1})} I did this question
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  2. #2
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    Re: have difficulties with Geometric series questions

    $\begin{align*}

    &\phantom{=}\displaystyle \sum_{k=0}^{n-1} \dfrac{1}{a r^k} \\ \\

    &=\dfrac 1 a \sum_{k=0}^{n-1}r^{-k}\\ \\

    &= \dfrac 1 a \dfrac{r^{-n}-1}{r^{-1}-1}\\ \\

    &=\dfrac 1 a \dfrac{r^{1-n} - r}{1-r} \\ \\

    &=\dfrac{r^{1-n}}{a}\dfrac{r^n-1}{r-1}

    \end{align*}$

    $\begin{align*}

    \displaystyle

    S_n &= a \sum_{k=0}^{n-1}r^k \\ \\

    &= a \dfrac{r^n-1}{r-1}

    \end{align*}$

    so

    $\displaystyle \sum_{k=0}^{n-1} \dfrac{1}{a r^k} = \dfrac{r^{1-n}S_n}{a^2}$
    Last edited by romsek; Apr 17th 2017 at 08:49 AM.
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  3. #3
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    Re: have difficulties with Geometric series questions

    Quote Originally Posted by romsek View Post
    $\
    Thank you very much for taking your time to help me with this question
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