(1)
n=1時,
2√S1=2√a1=a1+1
4a1=(a1+1)²
(a1-1)²=0
a1=1
n≥2時,
2√Sn=an +1
4Sn=(an +1)²
4Sn-1=[a(n-1)+1]²
4Sn-4Sn-1=4an=(an +1)²-[a(n-1)+1]²
(an -1)²-[a(n-1)+1]²=0
[an-1+a(n-1)+1][an -1-a(n-1)-1]=0
[an+a(n-1)][an-a(n-1)-2]=0
數(shù)列各項(xiàng)均為正,an+a(n-1)>0,要等式成立,則an-a(n-1)=2,為定值.
數(shù)列{an}是以1為首項(xiàng),2為公差的等差數(shù)列.
an=1+2(n-1)=2n-1
數(shù)列{an}的通項(xiàng)公式為an=2n-1.
(2)
bn=1/[ana(n+1)]=1/[(2n-1)(2n+1)]=(1/2)[1/(2n-1)-1/(2n+1)]
前n項(xiàng)和Tn=b1+b2+...+bn=(1/2)[1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)]
=(1/2)[1-1/(2n+1)]
=n/(2n+1)