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Showing a Sum

  1. May 28, 2006 #1
    I want to show that [tex] \sum_{k=0}^{n} x^{k} = \frac{1-x^{n+1}}{1-x} [/tex] using the additive, homogeneous, and telescoping properties of summation. In a hint it says to write the sum as [tex] (1-x)\sum_{k=0}^{n} x^{k} [/tex]. How did they arrive at this? Did they factor out the [tex] 1-x [/tex]. I dont see how they did this. I would then write [tex] x^{k} [/tex] as [tex] x^{k+1} - (x-1)^{k+1} [/tex]. Then what?

    Thanks
     
    Last edited: May 28, 2006
  2. jcsd
  3. May 28, 2006 #2

    Curious3141

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    Homework Helper

    Let the sum be represented by S. You want to find a neat expression for S.

    One way to do that is to multiply S by (1-x). List out a few terms of S (maybe the first four, then "..." then the last two terms). Now multiply that by (1-x) term by term. Add up the like powers (same exponent of x) and see what happens. You will end up very quickly at the required proof.
     
  4. May 28, 2006 #3

    VietDao29

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    Another way is that:
    Let:
    [tex]S = \sum_{n = 1} ^ k (x ^ n)[/tex]
    Now multiply both sides by x, we have:
    [tex]x S = \sum_{n = 1} ^ k (x ^ {n + 1}) = \sum_{n = 2} ^ {k + 1} (x ^ n)[/tex]
    Now, from the 2 equations above, what can you do to get (1 - x) S?
    Can you go from here? :)
     
  5. May 28, 2006 #4
    Yup, I got it.

    Thanks :smile:
     
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