Set [tex]n=1[/tex]. Then
[tex]\displaystyle\sum_{i=1}^1(3i-1)=3(1)-1=2[/tex]
[tex]\dfrac{1(3(1)+1)}2=\dfrac42=2[/tex]
Both sides match, so the statement holds for this case.
Assume it holds for [tex]n=k[/tex]. Then
[tex]\displaystyle\sum_{i=1}^{k+1}(3i-1)=\sum_{i=1}^k(3i-1)+3(k+1)-1[/tex]
[tex]=\dfrac{k(3k+1)}2+3(k+1)-1[/tex]
[tex]=\dfrac{k(3k+1)}2+\dfrac{2(3k+2)}2[/tex]
[tex]=\dfrac{3k^2+7k+4}2[/tex]
[tex]=\dfrac{(k+1)(3k+4)}2[/tex]
[tex]=\dfrac{(k+1)(3(k+1)+1)}2[/tex]
so that the statement also holds for [tex]n=k+1[/tex], thus proving the statement by induction.
