Example 17 Let the function $f(x)$ be defined on $\mathbf{R}$, for any $x \in \mathbf{R}$, we have $f(1+4 x)=f(3-2 x), f(2+$ $3 x)=-f(7-6 x)$, find the value of $\sum_{k=1}^{100}[f(2 k-1)+f(4 k-2)]$.
From the conditions, we have $f(1)=f(1+4 \cdot 0)=f(3-2 \cdot 0)=f(3)=f\left(2+3 \cdot \frac{1}{3}\right)=$ $-f\left(7-6 \cdot \frac{1}{3}\right)=-f(5)$, and $f(1)=f(3-2 \cdot 1)=f(1+4 \cdot 1)=f(5)$, thus $f(1)=f(3)=f(5)=0$.

Also, $f(2)=f\left(1+4 \cdot \frac{1}{4}\right)=f\left(3-2 \cdot \frac{1}{4}\right)=f\left(\frac{5}{2}\right)=f\left(2+3 \cdot \frac{1}{6}\right)=-f(7-$ $\left.6 \cdot \frac{1}{6}\right)=-f(6)$, hence $f(2)+f(6)=0$.

Furthermore, by conclusion 7(3), $f(x)$ is a periodic function, and $2\left|(2-3)+\frac{2}{4}(7-1)\right|=4$ is one of its positive periods.
$$
\begin{array}{l}
\text { Therefore, } \sum_{k=1}^{100}[f(2 k-1)+f(4 k-2)] \\
=\sum_{k=1}^{100} f(2 k-1)+\sum_{k=1}^{50}[f(8 k-2)+f(8 k-6)] \\
=\sum_{k=1}^{100} 0+\sum_{k=1}^{50} 0=0 .
\end{array}
$$