如果楼主的意思是 $c,a,c,b,c,a,c,b,\cdots$ 这种数列,那么其通项为\[\begin{split}f(n)&=\frac{c}{4}\sum_{k=0}^3e^{\frac{2k(n-1)\pi i}{4}}+\frac{a}{4}\sum_{k=0}^3e^{\frac{2k(n-2)\pi i}{4}}+\frac{c}{4}\sum_{k=0}^3e^{\frac{2k(n-3)\pi i}{4}}+\frac{b}{4}\sum_{k=0}^3e^{\frac{2kn\pi i}{4}}\\
&=\frac{1}{4} \left((-1)^n (a+b-2 c)-(a-b) \left(\cos\frac{n\pi}{2}+\cos \frac{3n\pi}{2}\right)+a+b+2 c\right)\end{split}\] |