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