1.
bn=2/(2an-1),b1=2/(2a1-1)=2,
an=1/2+1/bn,
a(n+1)=1-1/(4an),
1/2+1/b(n+1)=1-1/[4(1/2+1/bn)]
=1-1/(2+4/bn)
=1-bn/(4+2bn)
2/b(n+1)=1-bn/(2+bn)=2/(2+bn)
b(n+1)=2+bn
则bn为首项为2公差为2的等差数列；
bn=2+2(n-1)=2n
an=1/2+1/bn=1/2+1/(2n)=(1+1/n)/2.
2.
cn=2an/[(n+1)^2]=(1+1/n)/[(n+1)^2]
=[(n+1)/n]/[(n+1)^2]
=(1/n)/(n+1)
=1/[n(n+1)]
=1/n-1/(n+1)
sn=(1-1/2)+(1/2-1/3)+……+[1/n-1/(n+1)]
=1-1/(n+1)
=n/(n+1)