Answer
See explanations.
Work Step by Step
Step 1. Prove that the formula is true for $n=1$: $LHS=1$, $RHS=\frac{1(3-1)}{2}=1$, thus it is true for $n=1$
Step 2. Assume the formula is true for $n=k$: we have $1+4+7+...+(3k-2)=\frac{k(3k-1)}{2}$
Step 3. Prove that it is also true for $n=k+1$:
$LHS=1+4+7+...+(3k-2)+(3k+3-2)=\frac{k(3k-1)}{2}+(3k+1)=\frac{k(3k-1)+2(3k+1))}{2}=\frac{3k^2+5k+2}{2}$
$RHS=\frac{(k+1)(3k+3-1)}{2}=\frac{(k+1)(3k+2)}{2}=\frac{3k^2+5k+2}{2}=LHS$
Thus, the formula is also true for $n=k+1$
Step 4. With mathematical induction, we have proved that the formula is true for all natural numbers n.
