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Math Help - Unable to solve a simple AP-GP series :( ..please help..

  1. #1
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    Question Unable to solve a simple AP-GP series :( ..please help..

    Hello,

    Somehow i dont seem to be able to solve this simple looking series...pls help..

    Sum of ..... 1*(3^2) + 2(4^2) + 3(5^2) + 4(6^2) .....upto 15 terms

    So, we can write the Tn as n[(n+2)^2] ,

    Also
    note the following about the series : as the series is 9 , 32 , 75 , 144, 245...
    -- -- -- --
    23 43 69 101
    -- -- --
    20 26 32 .....
    Please help me get some ideas...

    Thanks..
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  2. #2
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    well i think i did not post a clear post..

    so as the series comes down to 9 , 32 , 75 , 144 , 245 , ...

    which obviously is not an AP... but the second order differences are in AP.

    32 -9 = 23
    75-32 = 43 43-23= 20
    144-75= 69 69-43=26
    245-144=101 101-69=32 ...

    Any help...??
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  3. #3
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    Hello, skyskiers

    1\cdot3^2 + 2\cdot4^2 + 3\cdot5^2  + 4\cdot6^2 + \cdots + 15\cdot17^2

    So, we can write: . T_k \:=\:k(k+2)^2 . . . . yes

    If you know some summation formulas, we can solve it this way . . .


    S_n \;=\;\sum^n_{k=1} k(k+2)^2

    . . =\;\sum^n_{k=1}\left(k^3 + 4k^2 + 4k\right)

    . . = \quad\sum^n_{k=1}k^3 \qquad + \qquad 4\sum^n_{k=1}k^2 \qquad + \qquad 4\sum^n_{k=1}k

    . . = \;\frac{n^2(n+1)^2}{4} + 4\cdot\frac{n(n+1)(2n+1)}{6} + 4\cdot\frac{n(n+1)}{2}

    . . which simplifies to: . S_n \;=\;\frac{n(n+1)(3n^2 + 19n + 32)}{12}


    Therefore: . S_{15} \;=\;\frac{15\cdot16\,(3\!\cdot\!15^2 + 19\!\cdot\!15 + 32)}{12} \;=\;19,840

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  4. #4
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    This is how dumb can i be...

    Thx for saving....Thanks a ton man!!
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