Question

« Medal! »
Find the sum of the series 3^(2) + 6^(2) + 9^(2) + . . . + 36^2.

Answer 1

\[\sum _{n=1}^{12} (3 n)^2=\]

Answer 2

\[\sum _{n=1}^{12} (3 n)^2=\sum _{n=1}^{12} 9n^2 = 9\sum _{n=1}^{12} n^2\]

Answer 3

No, it's not...just the terms listed above add up to 1422

Answer 4

The answer is 5850. I just have to show work

Answer 5

\[\sum n^2=\frac{ n \left( n+1 \right)\left( 2n+1 \right) }{ 6 }\]

Answer 6

a good site to use is wolf ram alpha

Answer 7

it gives you free step by step to any equation for free

Answer 8

No, not any longer...now you need a Pro subscription

Answer 9

you can get it for free

Answer 10

Okay what equation do I need tho use?

Answer 11

the equation @surjithayer gave you will give you the sum of \(n^2\), then multiply by 9 as I showed

Answer 12

put n=12

Answer 13

Thanks so much