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Math Help - Geometric Progression

  1. #1
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    Geometric Progression

    Hi... can anyone help me with this ?
    looks simple, i know! but i forgot the method =(

    basically i want to expand

    the sum of r between r = 1 upto n-1

    to get

    1/2n(n-1)

    how would i do this?


    thanks!
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  2. #2
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    Quote Originally Posted by matlabnoob View Post
    Hi... can anyone help me with this ?
    looks simple, i know! but i forgot the method =(

    basically i want to expand

    the sum of r between r = 1 upto n-1

    to get

    1/2n(n-1)

    how would i do this?


    thanks!
    Hi matlabnoob,

    notice that 1+2+3=(2-1)+2+(2+1)=2+2+2=3(2)

    and 1+2+3+4+5=(3-2)+(3-1)+3+(3+1)+(3+2)=3+3+3+3+3=5(3)

    Therefore, this is the same as (number\ of\ terms)(average\ value)

    For a large number of terms, simply take the average to be \frac{1st\ term+last\ term}{2}

    Therefore \sum_{r=1}^{n-1}r=\frac{1+n-1}{2}(n-1)=\frac{n(n-1)}{2}
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  3. #3
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    thank you for your help!

    i understood that. now im trying to do the same for ... r^2...

    and im getting the same answer as for r

    am i right to use the same method for r^2??


    thanks again!=]
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  4. #4
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    Doing what with r^2? You titled this "geometric progression" but the first example you gave was an arithmetic progression. What is it you are trying to do?
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  5. #5
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    Quote Originally Posted by matlabnoob View Post
    thank you for your help!

    i understood that. now im trying to do the same for ... r^2...

    and im getting the same answer as for r

    am i right to use the same method for r^2??


    thanks again!=]
    No, you cannot use the same method.
    The previous one shows how to develop a formula for summing consecutive natural numbers.

    If you want to sum consecutive squares, that's more involved.

    You would need to understand the previous one quite well before attempting
    to find a quick way to sum squares.

    Here is a way to do it using "telescoping".

    If you want to sum squares from r=1 to n-1, then you could use the following

    (2r+1)^3-(2r-1)^3=(2r+1)(2r+1)(2r+1)-(2r-1)(2r-1)(2r-1)

    =(4r^2+4r+1)(2r+1)-(4r^2-4r+1)(2r-1)= (8r^3+4r^2+8r^2+4r+2r+1)-(8r^3-4r^2-8r^2+4r+2r-1)

    =(8r^3+12r^2+6r+1)-(8r^3-12r^2+6r-1)=24r^2+2

    \Rightarrow\ (2r+1)^3-(2r-1)^3=24r^2+2

    Telescoping

    r=1.......\ (2r+1)^3-(2r-1)^3=3^3-1^3

    r=2.......\ (2r+1)^3-(2r-1)^3=5^3-3^3

    r=3.......\ (2r+1)^3-(2r-1)^3=7^3-5^3

    r=4.......\ (2r+1)^3-(2r-1)^3=9^3-7^3

    all the way to

    r=n-3....\ (2r+1)^3-(2r-1)^3=(2n-5)^3-(2n-7)^3

    r=n-2....\ (2r+1)^3-(2r-1)^3=(2n-3)^3-(2n-5)^3

    r=n-1....\ (2r+1)^3-(2r-1)^3=(2n-1)^3-(2n-3)^3

    When these are summed, all terms except (2n-1)^3 and 1 cancel.

    Hence \sum_{r=1}^{n-1}\left(24r^2+2\right)=(2n-1)^3-1

    2 summed n-1 times is 2(n-1)

    Hence

    24\sum_{r=1}^{n-1}r^2=(2n-1)^3-2(n-1)-1

    \sum_{r=1}^{n-1}r^2=\frac{(2n-1)^3-2(n-1)-1}{24}

    If we sum all the way to n, this is more commonly known as the sum of squares

    \sum_{r=1}^nr^2=\frac{n(n+1)(2n+1)}{6}
    Last edited by Archie Meade; March 17th 2010 at 04:59 AM.
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