17.
(1)
By mathematical induction, we can deduce that
$$a _ { n } = \begin{cases} \frac { 3 n - 1 } { 2 } & \text{if } 2 \nmid n \\ \frac { 2 n - 2 } { 2 } & \text{if } 2 \mid n \end{cases}$$
Thus $b _ { n } = a _ { 2 n } = 3 n - 1$ for $n \in \mathbb { Z } ^ { + }$, with $b _ { 1 } = 2, b _ { 2 } = 5$.
(2)
$$\begin{gathered} \sum _ { k = 1 } ^ { 20 } a _ { k } = \sum _ { k = 1 } ^ { 10 } a _ { 2 k - 1 } + \sum _ { k = 1 } ^ { 10 } a _ { 2 k } \\ = \sum _ { k = 1 } ^ { 10 } ( 3 k - 2 ) + \sum _ { k = 1 } ^ { 10 } ( 3 k - 1 ) \\ = 6 \sum _ { k = 1 } ^ { 10 } k - 30 = 300 \end{gathered}$$