每日一题_190926

数列\(\{a_n\}\)满足\(a_1=1\),\(a_{n+2}=a_{n+1}-a_n(n\in\mathbb{N}^\ast)\),则\(S_{2016}=\underline{\qquad\qquad}\).
解析: 记\(a_2=a\),则\(\{a_n\}\)中的各项依次为$$
1,a,a-1,-1,-a,1-a,1,a,\cdots$$
显然该数列为以\(6\)为周期的周期数列,从而$$
S_{2016}=0.$$