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Math Help - find sum

  1. #1
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    find sum

    (\frac{1}{3^3}+\frac{1}{3^4})+(\frac{2}{3^5}+\frac  {2}{3^6})+(\frac{3}{3^7}+\frac{3}{3^8})+(\frac{4}{  3^9}+\frac{4}{3^{10}})+.....
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by perash View Post
    (\frac{1}{3^3}+\frac{1}{3^4})+(\frac{2}{3^5}+\frac  {2}{3^6})+(\frac{3}{3^7}+\frac{3}{3^8})+(\frac{4}{  3^9}+\frac{4}{3^{10}})+.....
    Put each of the bracketed terms over a common denominator and simplify a bit. That will leave you with a series you should recognise and be able to sum.

    RonL
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  3. #3
    Bar0n janvdl's Avatar
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    Quote Originally Posted by perash View Post
    (\frac{1}{3^3}+\frac{1}{3^4})+(\frac{2}{3^5}+\frac  {2}{3^6})+(\frac{3}{3^7}+\frac{3}{3^8})+(\frac{4}{  3^9}+\frac{4}{3^{10}})+.....
    Off topic, I know, but this is a very cool series

    T_{n} = \frac{n}{3^{2n + 1}} + \frac{n}{3^{2n + 2}}
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Put each of the bracketed terms over a common denominator and simplify a bit. That will leave you with a series you should recognise and be able to sum.

    RonL
    i tried that and ended up with \frac 4{3^2} \sum_{n = 1}^{\infty} \frac n{9^n}. but i can't remember for the life of me how to sum something like that...
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Jhevon View Post
    i tried that and ended up with \frac 4{3^2} \sum_{n = 1}^{\infty} \frac n{9^n}. but i can't remember for the life of me how to sum something like that...
    Put:

    S(x)=\sum_{n=0}^{\infty} x^n=\frac{1}{1-x}


    Now differentiate term by term:

    S'(x)=\sum_{n=1}^{\infty} n x^{n-1}=\frac{1}{(1-x)^2}

    so:

    xS'(x)=\sum_{n=1}^{\infty} n x^n=\frac{x}{(1-x)^2}

    Now put x=1/9.

    (also see attachment)

    RonL
    Attached Thumbnails Attached Thumbnails find sum-gash.png  
    Last edited by CaptainBlack; January 4th 2008 at 03:54 AM.
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  6. #6
    Super Member wingless's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Put:
    S(x)=\sum_{n=1}^{\infty} x^n=\frac{1}{1-x}
    Hmm.. Wasn't that \frac{x}{1-x}?
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  7. #7
    Super Member PaulRS's Avatar
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    Without differentiating

    We want to find G(x)=\sum_{k=0}^{\infty}{\left(k\cdot{x^k}\right)} ( By the root test you can see that the series converges for |x|<1)

    Now we consider:
    x\cdot{G(x)}=\sum_{k=0}^{\infty}{\left(k\cdot{x^{k  +1}}\right)}=\sum_{k=1}^{\infty}{(k-1)\cdot{x^k}}

    So: G(x)-x\cdot{G(x)}=\sum_{k=0}^{\infty}{\left(k\cdot{x^k}  \right)}-\sum_{k=1}^{\infty}{(k-1)\cdot{x^k}}=\sum_{k=1}^{\infty}{(k-k+1)\cdot{x^k}}=\sum_{k=1}^{\infty}{x^k}=\frac{x}{  1-x} whenever |x|<1 (Because it is a geometric series)

    Thus: (1-x)\cdot{G(x)}=\frac{x}{1-x} and we obtain: G(x)=\frac{x}{(1-x)^2}
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  8. #8
    Super Member PaulRS's Avatar
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    Quote Originally Posted by wingless View Post
    Hmm.. Wasn't that \frac{x}{1-x}?
    You are right, but luckily we get the same result because the constant 1 vanishes when differentiating

    (1+x+x^2+...)'=(x+x^2+...)'
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  9. #9
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    Hello, perash!

    S \;=\;\left(\frac{1}{3^3}+\frac{1}{3^4}\right)+\lef  t(\frac{2}{3^5}+\frac{2}{3^6}\right)+\left(\frac{3  }{3^7}+\frac{3}{3^8}\right)+\left(\frac{4}{3^9}+\f  rac{4}{3^{10}}\right)+ \cdots

    We have: . . . . . . . . . . S \;=\;\frac{4}{3^4} + \frac{8}{3^6} + \frac{12}{3^8} + \frac{16}{3^{10}} + \cdots

    Factor: . . . . . . . . . . . S\;=\;\frac{4}{3^4}\left[1 + \frac{2}{3^2} + \frac{3}{3^4} + \frac{4}{3^4} + \cdots\right]\;\;{\color{blue}[1]}

    Multiply by \frac{1}{3^2}: . . . . . \frac{1}{9}S \;=\;\frac{4}{3^4}\left[\qquad \frac{1}{3^2} + \frac{2}{3^4} + \frac{3}{3^6} + \cdots\right]\;\;{\color{blue}[2]}

    Subtract [2] from [1]: . \frac{8}{9}S \;=\;\frac{4}{3^4}\underbrace{\left[1 + \frac{1}{3^2} + \frac{1}{3^4} + \frac{1}{3^6} + \cdots\right]}_{\text{geometric series: }a = 1,\:r = 1/9}


    Hence, we have: . . . . \frac{8}{9}S \;=\;\frac{4}{81}\left[\frac{9}{8}\right] \quad\Rightarrow\quad\boxed{S \:=\:\frac{1}{16}}

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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by wingless View Post
    Hmm.. Wasn't that \frac{x}{1-x}?
    No the lower limit of summation should have been 0, and now is (it is just a constant term that
    disappears on differentiation, or sometimes when just typing).

    RonL
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  11. #11
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    This wonderful solution


    thank you Soroban
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