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:   12 + 42 + 72 + . . . + (3n - 2)2 = n(6n^2-3n-1)/2

Answer:

[tex]S_1 = 12 \\ S_2 = 54 \\ S_3 = 126 \\ S_n = 15n^2 - 3n [/tex]

Explanation: 

[tex]S_1 = 12 \\ S_2 = 12 +42 = 54 \\ S_3 = 12 + 42 + 72 = 126[/tex]

To calculate for [tex]S_n[/tex], note that the series is an arithmetic series because each term is equal to the preceding term plus a constant, which is 30 (called common difference). 

We let

 [tex]a_1 = \text{ first term} = 12 \\ a_2 = \text{ second term} = 42 \\ a_3 = \text{ third term} = 72\\. \\. \\. \\a_n = n\text{th term}[/tex]

Since [tex]S_n[/tex] is an arithmetic series, [tex]a_1, a_2, a_3, ... , a_n[/tex] is an arithmetic sequence and so the formula for the nth term is given by

[tex]a_n = a_1 + d(n - 1) \\ a_n = 12 + 30(n - 1) \\ \boxed{a_n = 30n - 18}[/tex]

Where d = common difference = 30 

Now, since [tex]S_n[/tex] is an arithmetic series, we can use the formula for the sum of the arithmetic series, which is given by

[tex]S_n = \frac{n}{2} (a_1 + a_n) \\ \\ = \frac{n}{2} (12 + 30n - 18) \\ \\= \frac{n}{2} (30n - 6) \\ \\ = \frac{30n^2 - 6n}{2} \\ \\ \boxed{S_n = 15n^2 - 3n}[/tex]