## Precalculus (6th Edition) Blitzer

The given expression is: ${{a}_{1}}^{{}},{{a}_{2}}^{{}},{{a}_{3}}^{{}},{{a}_{4}}^{{}}\ldots,{{a}_{n}}\ldots$ So, ${{a}_{1}}$ could be anything either positive or negative. Similarly, to ${{a}_{1}}^{{}},{{a}_{2}}^{{}},{{a}_{3}}^{{}},{{a}_{4}}^{{}}\ldots,{{a}_{n}}\ldots$ The sequence is the set of positive integers which is dependent on the expression of ${{a}_{1}}^{{}},{{a}_{2}}^{{}},{{a}_{3}}^{{}},{{a}_{4}}^{{}}\ldots,{{a}_{n}}\ldots$ Therefore if ${{a}_{n}}=3{{n}^{2}}-{{n}^{3}}$ The first 2 terms are positive and the 3rd term is 0; after 4th term, all terms are negative. \begin{align} & {{a}_{1}}=3{{\left( 1 \right)}^{2}}-{{\left( 1 \right)}^{3}}=2 \\ & {{a}_{2}}=3{{\left( 2 \right)}^{2}}-{{\left( 2 \right)}^{3}}=4 \\ & {{a}_{3}}=3{{\left( 3 \right)}^{2}}-{{\left( 3 \right)}^{3}}=0 \\ & {{a}_{4}}=3{{\left( 4 \right)}^{2}}-{{\left( 4 \right)}^{3}}=-16 \\ & {{a}_{5}}=3{{\left( 5 \right)}^{2}}-{{\left( 5 \right)}^{3}}=-50 \\ \end{align}