27. 在数列{an}\{a_{n}\}{an}中,已知a1=1a_{1}=1a1=1,an+1−an=sin(n+1)π2{a_{n+1}}-{a_n}=\sin \frac{(n+1)\pi }{2}an+1−an=sin2(n+1)π,记SnS_{n}Sn为数列{an}\{a_{n}\}{an}的前nnn项和,则S2019=(S_{2019}=(S2019=( )))