设向量an=（cosnπ/6,sinnπ/6),n属于正整数,向量b(1,√3) 则y=|a1+b|^2+|a2+b|^2+```````+|a10+b|^2=?

ai^2=1(i=1,2,3,...,10)
b^2=4
∑(i=1 to 10)cos
=cos(π/6)+cos0+cos(π/6)+cos(2π/6)+...+cos(6π/6)+cos(5π/6)+cos(4π/6)
=cos(4π/6)
=-1/2
y=|a1+b|^2+|a2+b|^2+...+|a10+b|^2
=(a1^2+a2^2+...+a10^2)+2(a1*b+a2*b+...+a10*b)+10*b^2
=10+2*2[∑(i=1 to 10)cos]+40
=10+4*(-1/2)+40
=48

an=（cosnπ/6,sinnπ/6),
所以an^2= cos²nπ/6,sin²nπ/6=1.
b=(1,√3),所以b*2=4.
y=|a1+b|^2+|a2+b|^2+```````+|a10+b|^2
= a1^2+ a2^2+……+ a10^2+2a1b+2a2b+……+2a10b+10 b^2
= a1^2+ a2^2+……+ a10^2+2(a1+a2+……+a10)b+10 b^2
=10+2(a1+a2+……+a10)b+40
=50+2(a1+a2+……+a10)b
因为an=（cosnπ/6,sinnπ/6),
所以a1=(√3/2,1/2),
a2=(1/2,√3/2),a3=(0,1),a4=(-1/2,√3/2),
a5=(-√3/2,1/2),a6=(-1,0),a7=(-√3/2,-1/2),
a8=(-1/2,-√3/2),a9=(0,-1),a10=(1/2,-√3/2),
所以a1+a2+……+a10=(-1-√3/2,1/2),
(a1+a2+……+a10)b=(-1-√3/2,1/2)*(1,√3)
=-1-√3/2+√3/2=-1,
∴y=50-2=48.