Chapter 14 - Sequences, Series, and the Binomial Theorem - 14.1 Sequences and Series - 14.1 Exercise Set - Page 895: 88

$0,3,12,147,21,612$ and $467,078,547$.

${{a}_{n+1}}={{\left( {{a}_{n}} \right)}^{2}}+3$ ${{a}_{1}}=0$ $\begin{align} & {{a}_{n+1}}={{\left( {{a}_{n}} \right)}^{2}}+3 \\ & {{a}_{1+1}}={{\left( {{a}_{1}} \right)}^{2}}+3 \\ & {{a}_{2}}={{0}^{2}}+3 \\ & =3 \end{align}$ $\begin{align} & {{a}_{n+1}}={{\left( {{a}_{n}} \right)}^{2}}+3 \\ & {{a}_{2+1}}={{\left( {{a}_{2}} \right)}^{2}}+3 \\ & {{a}_{3}}={{3}^{2}}+3 \\ & =12 \end{align}$ $\begin{align} & {{a}_{n+1}}={{\left( {{a}_{n}} \right)}^{2}}+3 \\ & {{a}_{3+1}}={{\left( {{a}_{3}} \right)}^{2}}+3 \\ & {{a}_{4}}={{\left( 12 \right)}^{2}}+3 \\ & =147 \end{align}$ $\begin{align} & {{a}_{n+1}}={{\left( {{a}_{n}} \right)}^{2}}+3 \\ & {{a}_{4+1}}={{\left( {{a}_{4}} \right)}^{2}}+3 \\ & {{a}_{5}}={{\left( 144 \right)}^{2}}+3 \\ & =21,612 \end{align}$ $\begin{align} & {{a}_{n+1}}={{\left( {{a}_{n}} \right)}^{2}}+3 \\ & {{a}_{5+1}}={{\left( {{a}_{5}} \right)}^{2}}+3 \\ & {{a}_{6}}={{\left( 21,612 \right)}^{2}}+3 \\ & =467,078,547 \end{align}$ Thus, $\begin{align} & {{a}_{1}}=0 \\ & {{a}_{2}}=3 \\ & {{a}_{3}}=12 \\ \end{align}$ And, $\begin{align} & {{a}_{4}}=147 \\ & {{a}_{5}}=21,612 \\ & {{a}_{6}}=467,078,547 \\ \end{align}$ The first six terms are: $0,3,12,147,21,612$ and $467,078,547$.