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Thread: finite summation formula

  1. #1
    jut
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    finite summation formula

    $\displaystyle \sum_{k=1}^n a^k b^k=?$
    Last edited by jut; Sep 19th 2009 at 05:32 PM.
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    Hello, jut!

    Simplify: .$\displaystyle \sum_{k=1}^n a^k b^k$

    We have: .$\displaystyle S \;=\;ab + a^2b^2 + a^3b^3 + \hdots + a^nb^n$

    This is a geometric series with: .$\displaystyle \begin{Bmatrix}\text{first term: }ab \\ \text{common ratio: }ab \\ n\text{ terms} \end{Bmatrix}$

    Its sum is: .$\displaystyle S \;=\;ab\,\frac{(ab)^n - 1}{ab-1} $

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  3. #3
    jut
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    Wow, nice Soroban!
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  4. #4
    jut
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    Quote Originally Posted by Soroban View Post
    Hello, jut!

    Its sum is: .$\displaystyle S \;=\;ab\,\frac{(ab)^n - 1}{ab-1} $

    [/size]

    One more question:

    if a=e, and b=-2, I could plug that into the above result and it would be correct?
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    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by jut View Post
    One more question:

    if $\displaystyle a=\mathrm{e}$, and $\displaystyle b=-2$, I could plug that into the above result and it would be correct?
    sure, as long as $\displaystyle ab\neq1$, because it is a finite sum!
    In the case $\displaystyle ab=1$, you get $\displaystyle \sum_{k=1}^{n}ab=n$.
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