\(S_n = 1 + 4 + 1 +...+ 3n - 2\) \(\begin{pmatrix}\because \text{nth term of sequence is}\\a_n = a+ (n - 1)d\\= 1+(n-1)3\\= 3n- 2\end{pmatrix}\)
\(= \sum (3n - 2)\)
\(= 3\sum n - 2\sum 1\)
\(= \frac{3(n(n+1))}{2}-2n\)
\(= \frac{3n^2+ 3n - 4n}{2}\)
\(= \frac{3n^2 - n}{2}\)
\(= \frac{n(3n -1)}{2}\)
