If we have
[tex]\displaystyle S_n: \sum_{i=1}^n (3i-2)=\dfrac{n(3n-1)}{2}[/tex]
in order to write [tex]S_k[/tex] and [tex]S_{k+1}[/tex], we simply have to substitute "n" with "k" and "k+1", respectively.
So, we have
[tex]\displaystyle S_k: \sum_{i=1}^k (3i-2)=\dfrac{k(3k-1)}{2}[/tex]
[tex]\displaystyle S_{k+1}: \sum_{i=1}^{k+1} (3i-2)=\dfrac{(k+1)(3(k+1)-1)}{2}=\dfrac{(k+1)(3k+3-1)}{2}=\dfrac{(k+1)(3k+2)}{2}=\dfrac{3k^2+5k+2}{2}[/tex]
