如图，在平面直角坐标系$xOy$上，有一系列点$P_1(x_1,y_1)$，$P_2(x_2,y_2)$，$\cdots$，$P_n(x_n,y_n)$，每一个点$P_n(n \in \mathbf{N}^)$均位于抛物线$y = x^2(x \geq 0)$的图象上.点$F$为抛物线的焦点，以点$P_n$为圆心的$\odot P_n$都与$x$轴相切，且$\odot P_n$与$\odot P_{n+1}$外切.若$P_1F = \frac{5}{4}$，且$x_{n+1} < x_n(n \in \mathbf{N}^)$，$T_n = x_nx_{n+1}$，${T_n}$的前$n$项之和为$S_n$，则以下说法错误的是（ ）
A. $x_1 = 1$
B. $\left{\frac{1}{x_n}\right}$是等差数列
C. $S_8 = \frac{8}{17}$
D. $S_8 = \frac{16}{17}$
![图片[1]-抛物线+圆外切，构造等差，裂项求和](https://www.xuezizi.com/wp-content/uploads/2026/03/image-6.png)
抛物线+圆外切，构造等差，裂项求和
D
$\because |P_nP_{n+1}| = \sqrt{(x_{n+1} – x_n)^2 + (y_{n+1} – y_n)^2}$ $= \sqrt{(x_{n+1} – x_n)^2 + (x_{n+1}^2 – x_n^2)^2}$，
两圆半径之和为$y_n + y_{n+1} = x_n^2 + x_{n+1}^2$，$\therefore x_n^2 + x_{n+1}^2 = \sqrt{(x_{n+1} – x_n)^2 + (x_{n+1}^2 – x_n^2)^2}$$= (x_{n+1} – x_n)\sqrt{1 + (x_{n+1} + x_n)^2}$，
整理可得：$\left(\frac{1}{x_n} – \frac{1}{x_{n+1}}\right)^2 = 4$，$\because x_{n+1} < x_n$，$\therefore \frac{1}{x_{n+1}} – \frac{1}{x_n} = 2$，又$P_1F = x_1 + \frac{1}{4} = \frac{5}{4}$，得$\frac{1}{x_1} = 1$，
$\therefore$数列$\left{\frac{1}{x_n}\right}$是以$1$为首项，$2$为公差的等差数列，$\therefore \frac{1}{x_n} = 1 + 2(n – 1) = 2n – 1$，$\therefore x_n = \frac{1}{2n – 1}$，
$\therefore T_n = x_nx_{n+1} = \frac{1}{(2n – 1)(2n + 1)} = \frac{1}{2}\left(\frac{1}{2n – 1} – \frac{1}{2n + 1}\right)$，
$\therefore S_8 = \frac{1}{2}\left(1 – \frac{1}{3} + \frac{1}{3} – \frac{1}{5} + \cdots + \frac{1}{15} – \frac{1}{17}\right) $$= \frac{1}{2} \times \frac{16}{17} = \frac{8}{17}$．
故选：D．
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