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Geometric Sum

  1. Jun 2, 2008 #1
    1. The problem statement, all variables and given/known data

    Calculate the following sum:

    [tex]\sum_{k=1}^{10} 3 \cdot 2^{k}[/tex]

    3. The attempt at a solution

    [tex]\sum_{k=1}^{10} 3 \cdot 2^{k} = 3 \cdot \sum_{k=1}^{10} 2^{k} = 3 \cdot (2^{10} - 1)[/tex]

    Although it seems that I am missing something.
     
  2. jcsd
  3. Jun 2, 2008 #2

    D H

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    Very close but no cigar. How can [itex]\sum_{k=1}^{10} 2^{k} = 2^{10} - 1[/itex]? You are missing something from the right-hand side.
     
  4. Jun 3, 2008 #3
    Right, if you divide a term in the sum by the previous term, you get 2, not 1? So a factor of 2 is missing.
     
  5. Jun 3, 2008 #4

    Gib Z

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    Do you know [tex]x^n - y^n = (x-y)(x^{n-1} + x^{n-1}y + ... + y^{n-1} )[/tex] ?

    For (x,y) = (2,1) that forms into what you need. Look at the expansion of what you have, 2^10 - 1, and see how that falls different to the sum you want on the RHS, then change it accordingly =]
     
  6. Jun 3, 2008 #5
    **Edited by user.
     
    Last edited: Jun 3, 2008
  7. Jun 3, 2008 #6

    Gib Z

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    Forum regulations prohibit giving out answers to homework threads - please just give hints. Edit your post.
     
  8. Jun 3, 2008 #7
    ok sorry buddy!

    @author of thread... look up summation of a GP
     
  9. Jun 3, 2008 #8

    HallsofIvy

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    The formula you are trying to remember is
    [tex]\sum_{k=0}^n a r^k= a\frac{1-r^{n+1}}{1- r}[/itex]
    For r= 2, so that 1-r= 1-2= -1, that becomes
    [tex]\sum_{k=0}^n a 2^k= a (2^{n+1}- 1)[/tex]

    There are two important differences between that and the formula you are using.
     
  10. Jun 4, 2008 #9
    Let's see.

    The general formula is

    [tex]\sum_{k=0}^n a r^k= a\frac{1-r^{n+1}}{1- r}[/tex]

    Now the terms in the sum

    [tex]\sum_{k=1}^{10} 3 \cdot 2^{k} = 6 + 12 + 24 + ... + 3072[/tex]

    This means that

    n = 9 (10-1)
    r = 2 (taking the n:th term/(n-1):th term)
    a = 6 (the starting term)

    So

    [tex]\sum_{k=1}^{10} 3 \cdot 2^{k} = 6\frac{1-2^{9+1}}{1-2} = 6\frac{1-2^{10}}{1- 2} = 6(2^{10} - 1)[/tex]

    That sounds about right?

    Why is n = 9 here? Is that just the number of terms that have an exponent >= 2?
     
  11. Jun 5, 2008 #10
    Someone above told me not to give out the answer.. and btw your answer is correct.. and yes n = 10 not 9 . There are 10 terms of the GP.
     
  12. Jun 5, 2008 #11

    HallsofIvy

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    spdeyunlimit, n= 9, not 10 because because the sum starts with k= 1. He is thinking of this as 6+ 6(2)+ 6(22)+ 6(2n) with n running from 0 to 9.

    Another way to do the problem would be to use the same formula with n= 10, a= 3, but then subtract off 3- to allow for the missing 3(20) term.
     
  13. Jun 5, 2008 #12
    for the GP
    a, ar, ar^2, ar^3 ... and so on
    Sum of first n terms = a(1 - r^n) / (1-r)
    where in the first term the power of r is 0.. whereas here it is one.. so yeah i got that.. it will be for n = 9
     
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