Question

use the following rule to find 3^2 + 6^2 + 9^2 + .... + (3n)^2

Answer 1

\[1^2 + 2^2 + 3^2 + .... + n^2 = 1/6n(n+1)(2n+1)\]

Answer 2

take 3^2 common out.

Answer 3

no wait shouldn't i just replace n with 3n on the RHS of the rule??

Answer 4

That would have worked if you had
1^2 + 2^2 + 3^2 ....... (3n)^2

which is not the case here.

Answer 5

arrey but the first rule is what i posted, and the second one looks like thi,s if you couldn't make out
\[3^2 + 6^2 + 9^2 + .... + (3n)^2\]

Answer 6

I am not understand your confusion.

Answer 7

and what is 3^2 but (1*3)^2, and what is 6^2 but (2*3)^2

Answer 8

again, you can factor out 3^2 from each term
like 9^2 = 3^2.3^2
and you get 3^2(1^2+2^2+...)

Answer 9

(2*3)^2 = 2^2*3^2

Answer 10

yes, i know that :P

Answer 11

then your problem is done.

Answer 12

=3^2*(1/6)n(n+1)(2n+1)
=9*(1/6)n(n+1)(2n+1)
=(3/2)n(n+1)(2n+1)
got this ?