Answer
First five terms of the sequence are $\frac{1}{2},\frac{1}{5},\frac{1}{10},\frac{1}{17},\frac{1}{26}$, and the 12th term of the sequence is $\frac{1}{145}$.
Work Step by Step
${{a}_{n}}=\frac{1}{{{n}^{2}}+1}$
For the first term, substitute $n=1$ in the general term ${{a}_{n}}=\frac{1}{{{n}^{2}}+1}$,
$\begin{align}
& {{a}_{1}}=\frac{1}{{{1}^{2}}+1} \\
& =\frac{1}{2}
\end{align}$
For the second term, substitute $n=2$ in the general term ${{a}_{n}}=\frac{1}{{{n}^{2}}+1}$,
$\begin{align}
& {{a}_{2}}=\frac{1}{{{2}^{2}}+1} \\
& =\frac{1}{5}
\end{align}$
For the third term, substitute $n=3$ in the general term ${{a}_{n}}=\frac{1}{{{n}^{2}}+1}$,
$\begin{align}
& {{a}_{3}}=\frac{1}{{{3}^{2}}+1} \\
& =\frac{1}{10}
\end{align}$
For the fourth term, substitute $n=4$ in the general term ${{a}_{n}}=\frac{1}{{{n}^{2}}+1}$,
$\begin{align}
& {{a}_{4}}=\frac{1}{{{4}^{2}}+1} \\
& =\frac{1}{17}
\end{align}$
For the fifth term, substitute $n=5$ in the general term ${{a}_{n}}=\frac{1}{{{n}^{2}}+1}$,
$\begin{align}
& {{a}_{5}}=\frac{1}{{{5}^{2}}+1} \\
& =\frac{1}{26}
\end{align}$
For the 12th term in the sequence, substitute $n=12$ in the general term ${{a}_{n}}=\frac{1}{{{n}^{2}}+1}$
$\begin{align}
& {{a}_{12}}=\frac{1}{{{12}^{2}}+1} \\
& =\frac{1}{145}
\end{align}$
