Consecutive sum of exponentiations

  • #1
181
0
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!
 

Answers and Replies

  • #2
798
34
Simply expand :
(x^0+x^1+x^2+...+x^n)(x-1)
 
  • #3
181
0
I didn't understand why you wrote (x-1) after the expansion... thank you
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
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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].
 
  • #5
798
34
I didn't understand why you wrote (x-1) after the expansion... thank you
You will understand if you make the multiplication of the series by (x-1) and then simplify. Just do it !
 

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