Answer
$1,3,13,63,313,1563$.
Work Step by Step
${{a}_{n+1}}=5{{a}_{n}}-2$
${{a}_{1}}=1$
$\begin{align}
& {{a}_{n+1}}=5{{a}_{n}}-2 \\
& {{a}_{1+1}}=5{{a}_{1}}-2 \\
& {{a}_{2}}=5\cdot 1-2 \\
& =3
\end{align}$
$\begin{align}
& {{a}_{n+1}}=5{{a}_{n}}-2 \\
& {{a}_{2+1}}=5{{a}_{2}}-2 \\
& {{a}_{3}}=5\cdot 3-2 \\
& =13
\end{align}$
$\begin{align}
& {{a}_{n+1}}=5{{a}_{n}}-2 \\
& {{a}_{3+1}}=5{{a}_{3}}-2 \\
& {{a}_{4}}=5\cdot 13-2 \\
& =63
\end{align}$
$\begin{align}
& {{a}_{n+1}}=5{{a}_{n}}-2 \\
& {{a}_{4+1}}=5{{a}_{4}}-2 \\
& {{a}_{5}}=5\cdot 63-2 \\
& =313
\end{align}$
$\begin{align}
& {{a}_{n+1}}=5{{a}_{n}}-2 \\
& {{a}_{5+1}}=5{{a}_{5}}-2 \\
& {{a}_{6}}=5\cdot 313-2 \\
& =1563
\end{align}$
Therefore,
$\begin{align}
& {{a}_{1}}=1 \\
& {{a}_{2}}=3 \\
& {{a}_{3}}=13 \\
& \\
\end{align}$
And,
$\begin{align}
& {{a}_{4}}=63 \\
& {{a}_{5}}=313 \\
& {{a}_{6}}=1563 \\
\end{align}$
Thus, the first six terms are:
$1,3,13,63,313,1563$.