Answer
The first four terms are,${{a}_{1}}=0,{{a}_{2}}=\frac{1}{5},{{a}_{3}}=\frac{1}{5}\text{ and }{{a}_{4}}=\frac{3}{17}$
The $8\text{th}$ term is ${{a}_{8}}=\frac{7}{65}$, and the $12\text{th}$ term is ${{a}_{12}}=\frac{11}{145}$.
Work Step by Step
${{a}_{n}}=\frac{n-1}{{{n}^{2}}+1}$ …… (1)
To find the value of the first term, ${{a}_{1}}$ put $n=1$ in equation (1),
$\begin{align}
& {{a}_{n}}=\frac{n-1}{{{n}^{2}}+1} \\
& {{a}_{1}}=\frac{1-1}{{{1}^{2}}+1} \\
& =0
\end{align}$
For the second term, ${{a}_{2}}$ put $n=2$ in equation (1)
$\begin{align}
& {{a}_{n}}=\frac{n-1}{{{n}^{2}}+1} \\
& {{a}_{2}}=\frac{2-1}{{{2}^{2}}+1} \\
& =\frac{1}{5}
\end{align}$
For the third term, ${{a}_{3}}$ put $n=3$ in equation (1)
$\begin{align}
& {{a}_{n}}=\frac{n-1}{{{n}^{2}}+1} \\
& {{a}_{3}}=\frac{3-1}{{{3}^{2}}+1} \\
& =\frac{2}{10} \\
& =\frac{1}{5}
\end{align}$
For the fourth term, ${{a}_{4}}$ put $n=4$ in equation (1)
$\begin{align}
& {{a}_{n}}=\frac{n-1}{{{n}^{2}}+1} \\
& {{a}_{4}}=\frac{4-1}{{{4}^{2}}+1} \\
& =\frac{3}{17}
\end{align}$
For the $8\text{th}$ term, ${{a}_{8}}$ put $n=8$ in equation (1)
$\begin{align}
& {{a}_{n}}=\frac{n-1}{{{n}^{2}}+1} \\
& {{a}_{8}}=\frac{8-1}{{{8}^{2}}+1} \\
& =\frac{7}{65}
\end{align}$
For the $\text{12th}$ term, ${{a}_{12}}$ put $n=12$ in equation (1)
$\begin{align}
& {{a}_{n}}=\frac{n-1}{{{n}^{2}}+1} \\
& {{a}_{12}}=\frac{12-1}{{{12}^{2}}+1} \\
& =\frac{11}{144+1} \\
& =\frac{11}{145}
\end{align}$
