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Thread: How do you do this?

  1. #1
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    How do you do this?

    Hi, I've got some problems for homework and I'm not sure how to do this one:

    Sum the series:

    3^2 + 6^2 + 9^2 + ... + 2997^2 + 3000^2


    Could someone please explain how it's done?
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  2. #2
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    I think I got it.

    Is the answer 1001500500?
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by sstr
    Hi, I've got some problems for homework and I'm not sure how to do this one:

    Sum the series:

    3^2 + 6^2 + 9^2 + ... + 2997^2 + 3000^2


    Could someone please explain how it's done?
    $\displaystyle 3^2 + 6^2 + 9^2 + ... + 2997^2 + 3000^2=$
    $\displaystyle
    =9(1^2+2^2+..+999^2+1000^2)=$$\displaystyle \mathrm{XXXXXXXXXXXX}
    $

    CENSORED BY AUTHOR because this is one of the questions in an enrichment
    series (problem 5 in the take home Noether Student problems for the 2006
    Nerichment program run by the Australian MAthematical Olympiad Committee)

    RonL
    Last edited by CaptainBlack; Jul 10th 2006 at 05:00 AM.
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  4. #4
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    Sorry, I don't understand.

    Why is it 9 outside the brackets?
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  5. #5
    Super Member malaygoel's Avatar
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    Quote Originally Posted by sstr
    Sorry, I don't understand.

    Why is it 9 outside the brackets?
    The general term of the sequence $\displaystyle (3n)^2$ which is equivalent to $\displaystyle 9n^2$.

    Keep Smiling
    Malay
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  6. #6
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    Hello, sstr!

    Let me give it a try . . .


    Sum the series: .$\displaystyle 3^2 + 6^2 + 9^2 + 12^2 + \hdots + 2997^2 + 3000^2$

    We have: .$\displaystyle X \;= \;(3\!\cdot\!1)^2 + (3\!\cdot\!2)^2 + (3\!\cdot\!3)^2 + (3\!\cdot\!4)^2$$\displaystyle + \hdots + (3\!\cdot\!999)^2 + (3\!\cdot\!1000)^2$

    . . . . . . . .$\displaystyle X \;= \;3^2\!\cdot\!1^2 + 3^2\!\cdot\!2^2 + 3^2\!\cdot\!3^2 + 3^2\!\cdot\!4^2 + \hdots + 3^2\!\cdot\!999^2 + 3^2\!\cdot\!1000^2$

    Factor:. . .$\displaystyle X\;=\;3^2\underbrace{\left(1^2 + 2^2 + 3^2 + 4^2 + \hdots + 999^2 + 1000^2\right)}$
    . . . . . . . . . . . . This is the sum of the first 1000 squares


    The sum of the first $\displaystyle n$ squares is given by: .$\displaystyle S_n\;=\;\frac{n(n+1)(2n+1)}{6}$

    . . Hence: .$\displaystyle S_{1000}\;=\;\frac{1000\cdot1001\cdot2001}{6}\:=\; 333,833,500$


    Therefore: .$\displaystyle X\;=\;9 \times 333,833,500\quad\Rightarrow\quad\boxed{X\,=\,3,004 ,501,500}$

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  7. #7
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    Thank you everyone, I understand now.
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