数列an的通项公式an=n^2乘sin(n/2π-π/4)前n项和为Sn 则S40等于

let bn=sin(n/2π-π/4)=sin((2n-1)pi/4)
then we follows b(4k+1)=b(4k+2)=2^0.5/2,b(4k+3)=b(4k+4)=-2^0.5/2;k=0,1,...
so a(4k+1)=(4k+1)^2*b(4k+1)=(2^0.5/2)*(4k+1)^2=a(4k+2)
a(4k+3)=(4k+3)^2*b(4k+3)=(-2^0.5/2)*(4k+3)^2=a(4k+4);k=0,1,...
then Sn=sum(a(4k+1)+a(4k+3)+a(4k+2)+a(4k+4));k=0,1,...,
now we get
S40=sum((2^0.5/2)*((4k+1)^2-(4k+3)^2))+sum((2^0.5/2)*((4k+2)^2-(4k+4)^2))
=sum(-2^0.5(8k+4+8k+6))=-2^0.5sum(16k+10);k=0,1,...,9
hence
S40=-820(2^0.5)