急啊！数列an的前n项和Sn=π/12(2n^2+n)

求AN(我会）

bn=sinan乘以sinan+1乘以sinan+2,求证：bn是等比数列

Sn=π/12(2n^2+n)

S(n-1)=π/12(2(n-1)^2+n-1)

an=π/12(2(2n-1)+1)=π/12(4n-1)=πn/3-π/12

a(n+1)=πn/3+π/4

a(n+2)=πn/3+7π/12

bn=sin(πn/3-π/12)sin(πn/3+π/4)sin(πn/3+7π/12)

=-sin(πn/3+11π/12)sin(πn/3+π/4)sin(πn/3+7π/12)

b(n+1)=sin(πn/3+π/4)sin(πn/3+7π/12)sin(πn/3+11π/12)

b(n+2)=sin(πn/3+7π/12)sin(πn/3+11π/12)sin(πn/3+15π/12)

=-sin(πn/3+7π/12)sin(πn/3+11π/12)sin(πn/3+π/4)

b(n+1)*b(n+1)=[sin(πn/3+π/4)sin(πn/3+7π/12)sin(πn/3+11π/12)]^2=bn*b(n+2)

bn是等比数列