题目内容：

2.正项数列｛an｝前n项和为Sn,且2*根号Sn=an+1；（1）求数列{an}通项公式；（2）令bn=1/[an*（an+1)],设数

优质解答

(1)
n=1时,
2√S1=2√a1=a1+1
4a1=(a1+1)²
(a1-1)²=0
a1=1
n≥2时,
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
数列各项均为正,an+a(n-1)>0,要等式成立,则an-a(n-1)=2,为定值.
数列{an}是以1为首项,2为公差的等差数列.
an=1+2(n-1)=2n-1
数列{an}的通项公式为an=2n-1.
(2)
bn=1/[ana(n+1)]=1/[(2n-1)(2n+1)]=(1/2)[1/(2n-1)-1/(2n+1)]
前n项和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)