Question

Sn: 12 + 42 + 72 + . . . + (3n - 2)2 =

Answer 1

oops forgot to put the first part, sorry. A statement Sn about the positive integers is given. Write statements S1, S2, and S3, and show that each of these statements is true.

Also the 2s are meant to be exponents

Answer 2

\[
S_n = 1^2 + 4^2 + 7^2 + .... + (3n-2)^2 \\
S_n = \frac n2(6n^2-3n-1) \\
S_1 = \frac 12(6(1)^2-3(1)-1) = \frac 12 (6-3-1) = 1 \\
S_2 = ?
\]

Answer 3

just put n = 2 in the formula and calculate S2
then put n = 3 in the formula and calculate S3

where \(S_n\) is the sum of the first n terms of the series.
So if n = 1, we have just one term in the series which is 1^2 = 1.
In my previous reply I have plugged n = 1 into the formula and verified S1 = 1.

Next try n = 2 both in the series and in the formula and prove they are the same.

Then the same with n = 3.

Answer 4

okay, I think I got it :) thank you