一道高三数学题,高手入

已知函数$f(x)=\dfrac{1}{x^2-4x+3}$，则$f(4)+f(5)+f(6)+\cdots+f(2016)$的值为多少？

解法一：

将$f(x)$分解为$\dfrac{1}{(x-1)(x-3)}$，则原式可化为：

$$\begin{aligned} &f(4)+f(5)+f(6)+\cdots+f(2016) \\ =&\dfrac{1}{3\times1}+\dfrac{1}{3\times2}+\dfrac{1}{4\times2}+\dfrac{1}{4\times3}+\cdots+\dfrac{1}{2015\times2013}+\dfrac{1}{2015\times2016} \\ =&\left(\dfrac{1}{1}-\dfrac{1}{3}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\cdots+\left(\dfrac{1}{2013}-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2015}-\dfrac{1}{2016}\right) \\ =&1-\dfrac{1}{2016} \\ =&\dfrac{2015}{2016} \end{aligned}$$

解法二：

设$S=f(4)+f(5)+f(6)+\cdots+f(2016)$，则：

$$\begin{aligned} S&=\dfrac{1}{(4-1)(4-3)}+\dfrac{1}{(5-1)(5-3)}+\dfrac{1}{(6-1)(6-3)}+\cdots+\dfrac{1}{(2016-1)(2016-3)} \\ &=\dfrac{1}{3}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+\cdots+\dfrac{1}{2013}-\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2016}\right) \\ &=\dfrac{1}{3}\left(1-\dfrac{1}{2016}\right) \\ &=\dfrac{2015}{2016} \end{aligned}$$

综上，$f(4)+f(5)+f(6)+\cdots+f(2016)=\dfrac{2015}{2016}$。