# Thread: How do you do this?

1. ## 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?

2. I think I got it.

3. 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?
$3^2 + 6^2 + 9^2 + ... + 2997^2 + 3000^2=$
$
=9(1^2+2^2+..+999^2+1000^2)=$
$\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

4. Sorry, I don't understand.

Why is it 9 outside the brackets?

5. Originally Posted by sstr
Sorry, I don't understand.

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

Keep Smiling
Malay

6. Hello, sstr!

Let me give it a try . . .

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

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

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

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

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

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

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

7. Thank you everyone, I understand now.