Question

A statement Sn about the positive integers is given. Write statements S1, S2, and S3, and show that each of these statements is true. Show your work.

Sn: 1^2 + 4^2 + 7^2 + . . . + (3n - 2)^2 = (n(〖6n〗^2-3n-1))/2


S1:1^2+4^2+7^2+…+(3*1-2)^2=1(6*1^2-3n-1)/2
S1: 1^2 = 1 (6*1^2-3*1-1))/2
S2: 1^2 + 4^2 = 2 (6*2^2-3*2-1))/2
S3: 1^2 + 4^2 + 7^2 = 3 ( 6*3^2 - 3*3 -1 ) /2 ...

and S4:1^2+4^2+7^2=(3*4-2)^2=4(6*4^2-3*4-1)/2

Sn : 1^2 + 4^2 + 7^2 + ... + (3n-2) ^2 = n ( 6*n^2 - 3n - 1) / 2 is true for all n, positive integers n=1,2,3,...

Answer 1

You have Sn, but you have a typo, or the copy/paste mangled it. It should be
Sn: 1^2+...+(3n-2)^2 = n(6n^2-3n-1)/2
So,
S1: 1^2 = 1(6*1^2-3(1)-1)/2
S2: 1^2+4^2 = 2(6*2^2-3*2-1)/2
S3: 1^2+4^2+7^2 = 3(6*3^2-3*3-1)/2
which are all true

We have not shown it to be true for all n. To do that, you need to do a proof using induction.