an>0,2√Sn=an+1 求证√Sn是等差数列

4Sn=(an+1)^2
4S(n-1)=[a(n-1)+1]^2
an=Sn-S(n-1)
所以相减
4an=(an+1)^2-[a(n-1)+1]^2
(an+1)^2-4an=[a(n-1)+1]^2
(an-1)^2=[a(n-1)+1]^2
an-1=a(n-1)+1或an-1=-a(n-1)-1
an=a(n-1)+2或an=-a(n-1)
an>0
所以an=-a(n-1)不成立
所以an=a(n-1)+2
所以an是等差数列
S1=a1
所以2√a1=a1+1
(√a1-1)=0,an=1
an=2n-1
Sn=(1+2n-1)*n/2=n^2
√Sn=n
所以√Sn是等差数列

bn=1/(2n-1)
bnb(n+1)=1/(2n-1)(2n+1)=(1/2)[1/(2n-1)-1/(2n+1)]
所以前n项和
=(1/2)[1/1-1/3+1/3-1/5+……+1/(2n-1)-1/(2n+1)]
=(1/2)[1-1/(2n+1)]
=n/(2n+1)