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Math Help - sum a series - algebra

  1. #1
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    sum a series - algebra

    Find the sum of the series 2^2+4^2+6^2+...+50^2

    Using the above result , find the value of 1^2+3^2+5^2+...+49^2 .

    My working

     \sum^{25}_{r=1}4r^2=22100

    The question says use the above result and this is what i am confused with . How to use the result above to determine the sum of the following series . Is this the method of differences ? If yes , how do i apply ?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by thereddevils View Post
    Find the sum of the series 2^2+4^2+6^2+...+50^2

    Using the above result , find the value of 1^2+3^2+5^2+...+49^2 .

    My working

     \sum^{25}_{r=1}4r^2=22100
    nice

    The question says use the above result and this is what i am confused with . How to use the result above to determine the sum of the following series . Is this the method of differences ? If yes , how do i apply ?
    you know the formula for the sum of the first n squares. the first part is about the sum of the first even squares (that is, the squares of even integers), the second is asking for the sum of the first n odd squares. do you see what to do now? yes, finding a difference is involved
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  3. #3
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    Re:

     u_r=4r^2 but i don see that u_r=f(r)-f(r-1) , so how can i apply the method of differences?

    Thanks again ,Jhevon . 
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  4. #4
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    Quote Originally Posted by thereddevils View Post
     u_r=4r^2 but i don see that u_r=f(r)-f(r-1) , so how can i apply the method of differences?

    Thanks again ,Jhevon . 
    S = \sum_{r = 1}^{50} r^2 = 1^2 + 2^2 + 3^2 + \, .... \, + \, 49^2 + 50^2 can be calculated from the standard formula.

    But S = \sum_{r = 1}^{50} r^2 = (1^2 + 3^2 + \, .... \, + 49^2) +  (2^2 + 4^2 + \, .... \, + 50^2).

    Therefore (1^2 + 3^2 + \, .... \, + 49^2)  = \sum_{r = 1}^{50} r^2 - (2^2 + 4^2 + \, .... \, + 50^2) = S - 22100 = \, ....
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