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Geometric progression

  1. Jun 17, 2011 #1
    1. The problem statement, all variables and given/known data

    This isn't the whole question, I understand the prior parts but somehow stuck on the "easy" part :( Need to solve a geometric progession problem.. find the sum of:

    h(1 + 3^h + 3^2h + ... + 3^(n-1)h)

    Where nh = 1

    The sum should equal to (2h)/((3^h) -1) which is given as is a prove question

    2. Relevant equations

    N/A

    3. The attempt at a solution

    h(1 + 3^h + 3^2h + ... + 3^(n-1)h) --- this is the list

    (h(1-(3^h)^(n-1)))/(1-3^h) --- using the formula Sn = a(1-r^n)/(1-r)

    simplify slightly (using nh = 1)

    = (h(1-3^(1-h)))/(1-3^h)

    times top and bottom by -1 to match the denominator of answer

    = (h(3^(1-h) - 1))/((3^h) -1)

    and now I can't seem to get the final answer out.

    Hopefully I've included enough info for you guys to understand what I'm going on about lol - and apologise for the excess of brackets :S

    Thank you in advance :D
     
  2. jcsd
  3. Jun 17, 2011 #2

    micromass

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    Hi Maatttt0! :smile:

    That is not the correct formula, is it? Isn't the correct formula

    [tex]S_n=a\frac{1-r^{n+1}}{1-r}[/tex]

    so with an n+1 instead of an n...
     
  4. Jun 17, 2011 #3
    Advice: don't memorize any geometric series formula. Instead, try this:

    If you have a geometric progression in r, i.e. S = a + ar + ar2 + ar3 + ... + arn, compute S - rS = S(1 - r) and then re-arrange. This leads you exactly to the formula micromass posted, but it really helps you see what's going on. I prefer to memorize a technique than a formula.

    For your problem, you should start by re-writing S = h(1 + 3h + 32h + 33h + ... + 3(n-1)h) as S/h = 1 + 3h + 32h + ... + 3(n-1)h. Noting that this is a geometric progression in 3h, multiply both sides by 3h. Then, compute the difference (S/h) - (S/h)3h. Can you go from there?
     
  5. Jun 17, 2011 #4
    Hey guys,

    Thank you for both of the replies :)

    micromass; I'm didn't think it was n+1 :S my formula booklet just has n by itself.

    Unit; I got the answer to come out I think xD

    3[itex]^{h}[/itex] * [itex]\frac{S}{h}[/itex] = 3[itex]^{h}[/itex] + 3[itex]^{2h}[/itex] + ... + 3[itex]^{nh}[/itex]

    3[itex]^{h}[/itex] * [itex]\frac{S}{h}[/itex] = (3[itex]^{h}[/itex](1-3[itex]^{1}[/itex])) / (1-3[itex]^{h}[/itex])

    S = -2h / (1-3[itex]^{h}[/itex]) then just times top and bottom by -1.

    Is this what you were suggesting? Hinting to divide through by h and times through by 3^h was useful.
     
  6. Jun 17, 2011 #5
    Yes, that's exactly it! Well done :)
     
  7. Jun 17, 2011 #6
    Aha yay! Thank you Unit - muchly appreciated :D
     
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