已知等差数列an的前n项和为Sn,且对于任意的正整数n满足2根号下Sn=(an)+1

1.
2√Sn=an+1

4Sn=(an)^2+2an+1
4S1=(a1)^2+2a1+1=4a1,
a1=1

4S(n-1)=[a(n-1)]^2+2a(n-1)+1
4an=4[sn-s(n-1)]=(an)^2+2an-[a(n-1)]^2-2a(n-1)
(an)^2-2an-[a(n-1)]^2-2a(n-1)=0
[an+a(n-1)][an-a(n-1)]-2[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)-2]=0
an+a(n-1)=0或an-a(n-1)-2=0
an=a1(-1)^(n-1)=(-1)^(n-1)
或an=a1+2(n-1)=2n-1

经检验都符合题意
所以
an=(-1)^(n-1)或an=2n-1。

2.
bn=1/[ana(n+1)]

2-1,当an=(-1)^(n-1)时，
bn=1/[ana(n+1)]
=1/{[(-1)^(n-1)][(-1)^n]}
=1/[(-1)^(2n-1)]
=-1
Bn=-1*n=-n;

2-2,当an=2n-1时，
bn=1/[ana(n+1)]
=1/[(2n-1)(2n+1)]
=(1/2)[1/(2n-1)-1/(2n+1)]
2bn=1/(2n-1)-1/(2n+1)
2b(n-1)=1/(2n-3)-1/(2n-1)
2b(n-2)=1/(2n-5)-1/(2n-3)
2b(n-3)=1/(2n-7)-1/(2n-5)
……
2b3=1/5-1/7
2b2=1/3-1/5
2b1=1/1-1/3
两边相加：
2Bn=2[b1+b2+b3+……+b(n-3)+b(n-2)+b(n-1)+bn]
=1-1/(2n+1)
=2n/(2n+1)
Bn=n/(2n+1).

综上所述
an=(-1)^(n-1)时，Bn=-n
an=2n-1时，Bn=n/(2n+1).

bn=1/(2n+1)(2n+3)
=1/2[2/(2n+1)(2n+3)]
=(1/2)[(2n+3)-(2n+1)]/(2n+1)(2n+3)
=(1/2)[1/(2n+1)-1/(2n+3)]
所以Bn=(1/2)[1/3-1/5+1/5-1/7+……+1/(2n+1)-1/(2n+3)]
=(1/2)[1/3-1/(2n+3)]
=n/(6n+9)