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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- Jul 3rd 2006, 12:09 AM #1

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- Jul 3rd 2006, 12:38 AM #2

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- Jul 3rd 2006, 01:25 AM #3

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Originally Posted by**sstr**

$\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

- Jul 3rd 2006, 03:17 AM #4

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- Jul 3rd 2006, 03:25 AM #5

- Jul 3rd 2006, 09:55 AM #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: .$\displaystyle X\;=\;9 \times 333,833,500\quad\Rightarrow\quad\boxed{X\,=\,3,004 ,501,500}$

- Jul 3rd 2006, 11:55 PM #7

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