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Math Help - Proving a summation

  1. #1
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    Proving a summation

    Dear all, how can I prove the following summation (See attachment)?
    I tried to use integrals but the result was kind of broad.

    Attachment 29481
    Attached Thumbnails Attached Thumbnails Proving a summation-capture.jpg  
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  2. #2
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    Re: Proving a summation

    0 = \sum_{k = 0}^0 ka^k = \dfrac{a}{(1-a)^2}[1-(0+1)a^0 + 0a^{0+1}] = 0

    Assume the formula is correct for all nonnegative integers up to n.

    \sum_{k=0}^{n+1}ka^k = \sum_{k=0}^n ka^k + (n+1)a^{n+1}

    By the induction assumption, we have

    \sum_{k=0}^n ka^k + (n+1)a^{n+1} = \dfrac{a}{(1-a)^2}[1-(n+1)a^n + na^{n+1}] + (n+1)a^{n+1}

    Can you finish from here?
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  3. #3
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    Re: Proving a summation

    Alternately, if you start with

    \dfrac{1-a^{n+2}}{1-a} = \sum_{k=0}^{n+1}a^k

    And differentiate both sides with respect to a, you can also get the formula you want.
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  4. #4
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    Re: Proving a summation

    Thank you.
    abd Could you help me with how we can come up with the formula from the summation?
    As how to deduce the formula from the sigma notation?
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  5. #5
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    Re: Proving a summation

    The original formula? Or the formula I wrote: \dfrac{1-a^{n+2}}{1-a} = \sum_{k = 0}^{n+1}a^k? Because that formula is just the standard geometric sum:

    (1+a+a^2+\ldots+a^{n+1})(1-a)=
    \begin{array}{cccccccccccc} & 1 & + & a & + & a^2 & + & \ldots & + & a^{n+1} & & \\ & & - & a & - & a^2 & - & \ldots & - & a^{n+1} & - & a^{n+2} \\ = & 1 & & & & & & & & & - & a^{n+2}\end{array}
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  6. #6
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    Re: Proving a summation

    Imagine you have only the sigma notation, like this:
    \sum_{i = 0}^{n}i2^i = ?
    How would you deduce the following formula from the above summation?
    \frac{a}{(1 - a)^2}[1 - (n + 1)a^n + na^{n+1}]
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  7. #7
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    Re: Proving a summation

    I would think it looks like the derivative of a geometric sum. So, I would start with a geometric sum, and take the derivative to see what I get.
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