True
1) There are three steps, to using mathematical induction.
2) The first one is to check for n=1
[tex]\begin{gathered} \left(3n-2\right)^2=\frac{n\left(6n^2-3n-1\right)}{2} \\ \\ \left(3\cdot \:1-2\right)^2=\frac{1\cdot \left(6\cdot \:1^2-3\cdot \:1-1\right)}{2} \\ \\ (3-2)^2=\frac{1(6-3-1)}{2} \\ \\ 1^2=\frac{2}{2} \\ \\ 1=1\:\:True \end{gathered}[/tex]
3) Secondly, let's prove it for n=k:
[tex]\sum _{k=1}^k\left(3k-2\right)^2=\frac{k\left(6k^2-3k-1\right)}{2}[/tex]
4) And finally, for n=k+1
[tex]\begin{gathered} \sum _{n=1}^n\left(3n-2\right)^2=\frac{n\left(6n^2-3n-1\right)}{2} \\ \\ \sum _{\left(k+1\right)=1}^{k+1}\left(3\left(k+1\right)-2\right)^2=\frac{\left(k+1\right)\left(6\left(k+1\right)^2-3\left(k+1\right)-1\right)}{2} \\ \\ \left(3\left(k+1\right)-2\right)^2+\frac{k\left(6k^2-3k-1\right)}{2}=\frac{\left(k+1\right)\left(6\left(k+1\right)^2-3\left(k+1\right)-1\right)}{2} \\ \\ 2\left(3\left(k+1\right)-2\right)^2+k\left(6k^2-3k-1\right)=\left(k+1\right)\left(6\left(k+1\right)^2-3\left(k+1\right)-1\right) \\ \\ 6k^3+15k^2+11k+2=6k^3+15k^2+11k+2 \\ \\ 6k^3+15k^2+11k+2-2=6k^3+15k^2+11k+2-2 \\ \\ 6k^3+15k^2+11k=6k^3+15k^2+11k \\ \\ 6k^3+15k^2+11k-\left(6k^3+15k^2+11k\right)=6k^3+15k^2+11k-\left(6k^3+15k^2+11k\right) \\ \\ 0=0 \\ \\ True. \end{gathered}[/tex]
As all of those 3 steps were true. Then we can tell, that this is true.
