17. (10 points) Given the quadratic function
$$
f(x)=x^{2}-t x+t(t \in \mathbf{R})
$$

satisfies:
(1) The solution set of the inequality $f(x) \leqslant 0$ has exactly one element; $\square$
(2) If $0<x_{2}<x_{1}$, the inequality $f\left(x_{2}\right)<$ $f\left(x_{1}\right)$ always holds. Let the sequence $\left\{a_{n}\right\}$ have the sum of the first $n$ terms $S_{n}=f(n)$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Find $\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{a_{k} a_{k+2}}$.
Three, 17. (1) From condition (1), we have
$\Delta=t^{2}-4 t=0 \Rightarrow t=0$ or 4.
From condition (2), we know that $f(x)$ is an increasing function on $(0,+\infty)$. Therefore,
$$
t=0, f(x)=x^{2}, S_{n}=f(n)=n^{2}.
$$

Thus, $a_{n}=S_{n}-S_{n-1}=2 n-1(n \geqslant 2)$.
Also, $a_{1}=1$ fits, so $a_{n}=2 n-1$.
(2) Note that,
$$
\frac{1}{a_{n} a_{n+2}}=\frac{1}{4}\left(\frac{1}{a_{n}}-\frac{1}{a_{n+2}}\right).
$$

Then $\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{a_{k} a_{k+2}}$
$$
\begin{array}{l}
=\frac{1}{4} \lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\frac{1}{a_{k}}-\frac{1}{a_{k+2}}\right) \\
=\frac{1}{4} \lim _{n \rightarrow \infty}\left(1+\frac{1}{3}-\frac{1}{2 n+1}-\frac{1}{2 n+3}\right)=\frac{1}{3}.
\end{array}
$$