## Precalculus (6th Edition) Blitzer

$\underline{{{a}_{2}}=\frac{{{\left( -1 \right)}^{2}}}{{{4}^{2}}-1}=\frac{1}{15}}$.
Put $n=2$ to get the second term of the sequence given. Now, to get the third term, put $n=3$ and so on. We have the general form of the sequence: ${{a}_{n}}=\frac{{{\left( -1 \right)}^{n}}}{{{4}^{n}}-1}$. So to find the second term, put $n=2$: \begin{align} & {{a}_{2}}=\frac{{{\left( -1 \right)}^{2}}}{{{4}^{2}}-1} \\ & =\frac{1}{16-1} \\ & =\frac{1}{15} \end{align}