17. (12 points) In the sequence $\left\{a_{n}\right\}$, $a_{1}=1$, and its first $n$ terms sum $S_{n}$ satisfies the relation
$$
3 t S_{n}-(2 t+3) S_{n-1}=3 t(t>0, n=2,3, \cdots) \text {. }
$$
(1) Prove that the sequence $\left\{a_{n}\right\}$ is a geometric sequence;
(2) Let the common ratio of the sequence $\left\{a_{n}\right\}$ be $f(t)$, and construct the sequence $\left\{b_{n}\right\}$ such that $b_{1}=1, b_{n}=f\left(\frac{1}{b_{n-1}}\right)(n=2,3, \cdots)$, find $b_{n}$;
(3) Find the value of $b_{1} b_{2}-b_{2} b_{3}+b_{3} b_{4}-b_{4} b_{5}+\cdots+b_{2 n-1} b_{2 n}-b_{2 n} b_{2 n+1}$.
17. (1) From the given $3 S_{2}-(2 t+3) S_{1}=3 t$,

i.e., $3 t\left(a_{1}+a_{2}\right)-(2 t+3) a_{1}=3 t$.
Given $a_{1}=1$, solving for $a_{2}$ yields $a_{2}=\frac{2 t+3}{3 t}$.
Thus, $\frac{a_{2}}{a_{1}}=\frac{2 t+3}{3 t}$.
For $n \geqslant 2$, we have
$$
\begin{array}{l}
3 t S_{n+1}-(2 t+3) S_{n}=3 t, \\
3 t S_{n}-(2 t+3) S_{n-1}=3 t .
\end{array}
$$
(1) - (2) gives $3 t a_{n+1}-(2 t+3) a_{n}=0$.

Thus, $\frac{a_{n+1}}{a_{n}}=\frac{2 t+3}{3 t}$.
In summary, $\frac{a_{n+1}}{a_{n}}=\frac{2 t+3}{3 t}(n \geqslant 1)$.
Therefore, $\left\{a_{n}\right\}$ is a geometric sequence.
(2) From (1), we know $f(t)=\frac{2 t+3}{3 t}$, then
$$
b_{1}=1, b_{n}=\frac{2 \cdot \frac{1}{b_{n-1}}+3}{3 \cdot \frac{1}{b_{n-1}}}=\frac{2}{3}+b_{n-1} .
$$

Thus, $b_{n}-b_{n-1}=\frac{2}{3}(n=2,3, \cdots)$.
Therefore, $\left\{b_{n}\right\}$ is an arithmetic sequence, and
$$
b_{1}=1, d=b_{n}-b_{n-1}=\frac{2}{3} \text {. }
$$

Hence, $b_{n}=b_{1}+(n-1) d=\frac{2}{3} n+\frac{1}{3}$.
$$
\begin{array}{l}
\text { (3) } b_{1} b_{2}-b_{2} b_{3}+b_{3} b_{4}-b_{4} b_{5}+\cdots+b_{2 n-1} b_{2 n}-b_{2 n} b_{2 n+1} \\
=b_{2}\left(b_{1}-b_{3}\right)+b_{4}\left(b_{3}-b_{5}\right)+\cdots+b_{2 n}\left(b_{2 n-1}-b_{2 n+1}\right) \\
=-\frac{4}{3}\left(b_{2}+b_{4}+\cdots+b_{2 n}\right)=-\frac{4}{3} \cdot \frac{n\left(b_{2}+b_{2 n}\right)}{2} \\
=-\frac{4}{3} \cdot \frac{n\left(\frac{5}{3}+\frac{4 n+1}{3}\right)}{2}=-\frac{8}{9} n^{2}-\frac{4}{3} n .
\end{array}
$$