數列{an}的通項公式an=ncosnπ2+1,前n項和為Sn,則S2012=______.

數列{an}的通項公式an=ncosnπ2+1,前n項和為Sn,則S2012=______.

因為cosnπ2=0,-1,0,1,0,-1,0,1…;∴ncosnπ2=0,-2,0,4,0,-6,0,8…;∴ncosnπ2的每四項和為2;∴數列{an}的每四項和為:2+4=6.而2012÷4=503;∴S2012=503×6=3018.故答案為 ; ; ;3018.