Answer
See below.
Work Step by Step
1. Test for $n=1$, we have $LHS=1^2=1$, $RHS=\frac{1}{2}(1)(6-3-1)=1$, thus it is true for $n=1$.
2. Assume the statement is true for $n=k$, that is $1^2+4^2+7^2+...+(3k-2)^2=\frac{1}{2}(k)(6k^2-3k-1)$
3. For $n=k+1$, we have $LHS=1^2+4^2+7^2+...+(3k-2)^2+(3(k+1)-2)^2=\frac{1}{2}(k)(6k^2-3k-1)+(3k+1)^2=\frac{1}{2}(6k^3-3k^2-k+18k^2+12k+2)=\frac{1}{2}(6k^3+15k^2+11k+2)$. while $RHS=\frac{1}{2}(k+1)(6(k+1)^2-3(k+1)-1)=\frac{1}{2}(k+1)(6k^2+12k+6-3k-3-1)=\frac{1}{2}(k+1)(6k^2+9k+2)=\frac{1}{2}(6k^3+15k^2+11k+2)=LHS$
4. Thus the statement is true for all natural number of $n$.
