Given the parabola $y=-2 x^{2}+x-\frac{1}{8}$ and the point $A\left(\frac{1}{4}, \frac{11}{8}\right)$. Draw any line through the point $F\left(\frac{1}{4},-\frac{1}{8}\right)$, intersecting the parabola at points $B$ and $C$.
(1) Find the equation of the centroid trajectory of $\triangle A B C$, and express it in the form $y=f(x)$;
(2) If the sequence $\left\{x_{k}\right\}, 0<x_{1}<\frac{1}{2}$, satisfies $x_{k+1}=f\left(x_{k}\right)$. Prove: $\sum_{k=1}^{n} x_{k+1}^{k}<\frac{3}{5}$.
(1) Let the equation of the line passing through $F\left(\frac{1}{4},-\frac{1}{8}\right)$ be $y+\frac{1}{8}=k\left(x-\frac{1}{4}\right)$. Suppose $B\left(x_{1}, y_{1}\right), C\left(x_{2}, y_{2}\right)$, and solve the system of equations $\left\{\begin{array}{l}y+\frac{1}{8}=k\left(x-\frac{1}{4}\right), \\ y=-2 x^{2}+x-\frac{1}{8},\end{array}\right.$ to eliminate $y$, resulting in $2 x^{2}+(k-1) x-\frac{k}{4}=0$. Therefore, we have
$$
x_{1}+x_{2}=-\frac{k-1}{2}, y_{1}+y_{2}=k\left(x_{1}+x_{2}-\frac{1}{2}\right)-\frac{1}{4}=-\frac{k^{2}}{2}-\frac{1}{4} \text {. }
$$
$\left\{\begin{array}{l}x=\frac{3-2 k}{12}, \\ y=-\frac{k^{2}}{6}+\frac{3}{8},\end{array}\right.$ Eliminating $k$, we get $y=-6 x^{2}+3 x$.
(2) Since $0<x_{1}<\frac{1}{2}, x_{2}=f\left(x_{1}\right)=-6 x_{1}^{2}+3 x_{1}=3 x_{1}\left(1-2 x_{1}\right)$, we have
$$
0<x_{2} \doteq 3 x_{1}\left(1-2 x_{1}\right) \leqslant \frac{3}{2}\left(\frac{2 x_{1}+\left(1-2 x_{1}\right)}{2}\right)^{2}=\frac{3}{8},
$$

with equality holding if and only if $x_{1}=\frac{1}{4}$. Assuming $0<x_{k} \leqslant \frac{3}{8}$, then
$$
0<x_{k+1}=3 x_{k}\left(1-2 x_{k}\right) \leqslant \frac{3}{2}\left(\frac{2 x_{k}+\left(1-2 x_{k}\right)}{2}\right)^{2}=\frac{3}{8},
$$

with equality holding if and only if $x_{k}=\frac{1}{4}$. This implies $0<x_{k} \leqslant \frac{3}{8}(k=2,3, \cdots)$. Therefore,
$$
0<\sum_{k=1}^{n} x_{k+1}^{k} \leqslant \sum_{k=1}^{n}\left(\frac{3}{8}\right)^{k}=\frac{3}{5}\left[1-\left(\frac{3}{8}\right)^{n}\right]<\frac{3}{5} .
$$