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Consecutive sum of exponentiations

  1. Nov 22, 2011 #1
    Hello, i read that a sum of exponentiations like [itex]x^0+x^1+x^2+x^3...+x^n[/itex] can be solved with this forumula [itex]\frac{x^(n+1)-1}{x-1}[/itex], how is it possible do demonstrate this resolutive formula?

    Thank you!
     
  2. jcsd
  3. Nov 22, 2011 #2
    Simply expand :
    (x^0+x^1+x^2+...+x^n)(x-1)
     
  4. Nov 22, 2011 #3
    I didn't understand why you wrote (x-1) after the expansion... thank you
     
  5. Nov 22, 2011 #4

    HallsofIvy

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    Because of the x- 1 in the denominator in your equation.
    If A= B/(x- 1) then A(x- 1)= B.

    Is that true here?
    x
    That is not, however, how I would handle this problem. I would note that [itex]x^0+ x^1+ x^2+ \cdot\cdot\cdot+ x^n[/itex] is a geometric series with "common factor" x. The sum of a finite geometric series is
    [tex]\frac{1- x^{n+1}}{1- x}= \frac{x^{n+1}- 1}{x- 1}[/tex].
     
  6. Nov 22, 2011 #5
    You will understand if you make the multiplication of the series by (x-1) and then simplify. Just do it !
     
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