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Math Help - find the sum

  1. #1
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    find the sum

    find the sum of the given series:

    heres the equation: http://img19.imageshack.us/img19/559/untitledyc9.jpg

    I was told it could be done by first finding a and b using that 2n +1 = a(n+1) + b

    and given as a hint that E(n + 1)x^n is the derivateive of a known power series already.

    please help me with this.
    thankyou.
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  2. #2
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    Find the interval of absolute convergence and then.
    \sum_{n=0}^{\infty}(2n+1)x^n
    Expand,
    1+3x+5x^2+7x^3+...
    (2x+4x^2+6x^3+...)+(1+x+x^2+x^3+...)
    2x(1+2x+3x^2+...)+(1+x+x^2+...). (*)
    Okay, now we notice a very important thing.

    First we know that,
    1+x+x^2+x^3+...=\frac{1}{1-x}
    Then, take derivative,
    1+2x^2+3x^2+...=\frac{x}{(1-x)^2}.
    Substitute that into (*),
    2x\cdot \frac{x}{(1-x)^2}+\frac{1}{1-x}
    \frac{2x^2}{(1-x)^2}+\frac{1-x}{(1-x)^2}
    \frac{2x^2-x+1}{(1-x)^2}
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  3. #3
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    thanks for the help,

    but where does (2x + 4x^2 + 6x^3 +...) + (1 +...

    come from?


    also, what is the sum?

    thankyou.
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  4. #4
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    Quote Originally Posted by rcmango View Post
    thanks for the help,

    but where does (2x + 4x^2 + 6x^3 +...) + (1 +...

    come from?


    also, what is the sum?

    thankyou.
    Well, add them you get the same thing. No?
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