Answer
See explanations.
Work Step by Step
Step 1. Prove that the statement is true for $n=1$: $a_1=2\times3^1-2=4$, thus it is true for $n=1$
Step 2. Assume the statement is true for $n=k$: we have $a_k=2\times3^k-2$
Step 3. Prove that it is also true for $n=k+1$:
$a_{k+1}=3a_k+4=3(2\times3^k-2)+4=2\times3^{k+1}-6+4=2\times3^{k+1}-2$
Thus, the statement is also true for $n=k+1$
Step 4. With mathematical induction, we have proved that the statement is true for all natural numbers n.
