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
經(jīng)檢驗(yàn)都符合題意
所以
an=(-1)^(n-1)或an=2n-1.
2.
bn=1/[ana(n+1)]
2-1,當(dāng)an=(-1)^(n-1)時,
bn=1/[ana(n+1)]
=1/{[(-1)^(n-1)][(-1)^n]}
=1/[(-1)^(2n-1)]
=-1
Bn=-1*n=-n;
2-2,當(dāng)an=2n-1時,
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
兩邊相加：
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).
綜上所述
an=(-1)^(n-1)時,Bn=-n
an=2n-1時,Bn=n/(2n+1).