已知数列{an}的前n项和为Sn,且满足an+2Sn*Sn-1=0,a1=1/2.

证：
an+2SnSn-1=0
Sn-Sn-1+2SnSn-1=0
等式两边同除以SnSn-1
1/Sn-1-1/Sn+2=0
1/Sn-1/Sn-1=2,为定值.
1/S1=1/a1=2
数列{1/Sn}是以2为首项,2为公差的等差数列.
1/Sn=2+2(n-1)=2n Sn=1/(2n)
1/Sn-1=2+2(n-2)=2(n-1) Sn-1=1/[2(n-1)]
an=Sn-Sn-1=1/(2n)-1/[2(n-1)]=1/[2n(1-n)]
bn=2(1-n)an=1/n
b2²+b3²+...+bn²
=1/2²+1/3²+...+1/n²