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Geometric progression, is this correct? ratio/first/next term

  1. Nov 27, 2006 #1
    Hello everyone, i'm trying to solve the following:
    If n is an integer and n > = 1, find a formula for the expression
    [tex]2^n - 2^{n-1} + 2^{n-2} - 2^{n-3} + ... + (-1)^{n-1} * 2 + (-1)^n [/tex]

    okay this confuses me, because i'm not sure which is the first term and which is the last term...
    I figured the ratio was the following: -2
    becuase if u take 2^n/[-2^(n-1)] = -2
    I took the first term and divided it by the 2nd term to find the ratio or is it vice versa? by taking the 2nd term and dividing it by the first, which would give the ratio of: -1/2?

    once i find the ratio, i can find the term right after the last by multiplying the ratio by the last term in this case (-1)^n is the last term, and the first term is 2^n correct?

    Thanks, once i get this figured out i cna find the formual for the sum.
  2. jcsd
  3. Nov 27, 2006 #2
    You take the 2nd term and divide it by the first, so [tex] r = -\frac{1}{2} [/tex]
  4. Nov 27, 2006 #3
    Thanks courtrigrad
    Okay so the answer i'm getting would be:
    ["Mythical Next term" - "First real term" ]/[ratio - 1]
    thats the general formula to find the formula for the sum of a geometric progression and i come up with:

    sum = [-.5*(-1)^n - 2^n]/(-3/2)

    does that look correct to you? I'm not sure how i can clean that up though.
  5. Nov 27, 2006 #4
    [tex] \sum_{n=0}^{\infty} ar^{k} = \frac{a(1-r^{n+1})}{1-r} [/tex]

    The last term is [tex] (-1)^{n-1} 2^{n} [/tex]

    [tex] \frac{2^{n}(1-(-\frac{1}{2})^{n+1}))}{\frac{3}{2}} [/tex]
    Last edited: Nov 27, 2006
  6. Nov 27, 2006 #5
    thanks a ton! I now see that the (-1)^n was just changing the + to - and vice versa! :)
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