Write the first four terms of the sequence $\left\{a_{n}\right\}$ defined by the following recurrence relation.
\[
a_{n+1}=3 a_{n}^{2}+2 ; a_{1}=0
\]
\[
a_{1}=\square \text { (Simplify your answer.) }
\]
Answer
\(\boxed{\text{The first four terms of the sequence are } a_{1}=0, a_{2}=2, a_{3}=14, \text{ and } a_{4}=590}\)
Step 1 ：The problem asks for the first four terms of the sequence defined by the recurrence relation \(a_{n+1}=3 a_{n}^{2}+2\) with \(a_{1}=0\).
Step 2 ：To find these terms, we can start with \(a_{1}\) and use the recurrence relation to find \(a_{2}\), \(a_{3}\), and \(a_{4}\).
Step 3 ：Using the recurrence relation, we find that \(a_{2}=3*0^{2}+2=2\), \(a_{3}=3*2^{2}+2=14\), and \(a_{4}=3*14^{2}+2=590\).
Step 4 ：\(\boxed{\text{The first four terms of the sequence are } a_{1}=0, a_{2}=2, a_{3}=14, \text{ and } a_{4}=590}\)
