Answer
See below.
Work Step by Step
1. Test for $n=1$, we have $LHS=2$, $RHS=3^1-1=2$, thus it is true for $n=1$.
2. Assume the statement is true for $n=k$, that is $2+6+18+...+2\cdot3^{k-1}=3^k-1$
3. For $n=k+1$, we have $LHS=2+6+18+...+2\cdot3^{k-1}+2\cdot3^{k}=3^k-1+2\cdot3^{k}=3^{k+1}-1$
4. Thus the statement is true for all natural number of $n$.
