已知$f(x)=\sum\limits_{k=1}^{2017}\dfrac{\cos kx}{\cos^k x},$則$f(\dfrac{\pi}{2018})=$_____

分析:
設$g(x)=\sum\limits_{k=1}^{2017}\left(\dfrac{\cos kx}{\cos^k x}+i\dfrac{\sin kx}{\cos^k x}\right)$
$=\sum\limits_{k=1}^{2017}\left(\dfrac{\cos x+i\sin x}{\cos x}\right)^{k}$ 
$ =\dfrac{\frac{\cos x+i\sin x}{\cos x}-(\frac{\cos x+i\sin x}{\cos x})^{2018}}{1-\frac{\cos x+i\sin x}{\cos x}}$ 
$=\dfrac{\cos x+i\sin x-\frac{\cos2018x+i\sin2018x}{\cos^{2017}{x}}}{-i\sin x}$
則$g(\dfrac{\pi}{2018})=-1+i\dfrac{\cos\frac{\pi}{2018}+\cos^{-2017}{\frac{\pi}{2018}}}{\sin\frac{\pi}{2018}}$
比較實部可知$f(\dfrac{\pi}{2018})=-1$

思路參考：MT【34】