等差数列{an}的前n项和为Sn,已知(a2-1)^3=sin7/2π,(a2012-1)^3=sin9/2π,则s20

设首项为a1,公差为d,则
a2=a1+d,a2012=a1+2011d
sin(7π/2)=sin(-π/2+4π)=-1
sin(9π/2)=sin(π/2+4π)=1
∴(a2-1)^3=-1 => a2-1=-1 => a2=0
(a2012-1)^3=1 => a2012-1=1 => a2012=2
∴a1+d=0,a1+2011d=2
=> 2010d=2
∴S2013=(a1+a2013)*2013/2
=(2a1+2012d)*2013/2
=2010d*2013/2
=2*2013/2
=2013