一道数学题：已知数列an的前n项和为sn,满足an+2sns(n-1)=0(n≥2,n为正整数),a1=1/2

(1)an+2sns(n-1)=0,an=sn-s(n-1)化为1/sn-1/s(n-1)=2数列1/sn为等差数列。公差d=2,s1=a1=1/2,有1/s1=2,所以1/sn=（n-1)*d+1/s1=2nsn=1/(2n)an=sn-s(n-1)=1/[2n(1-n)](2)bn=2(1-n)an=2(1-n)/[2n(1-n)]=1/n,b2^2+b3^2+。+bn^2=(1/2)^2+(1/3)^2+。+(1/n)^2