Question

正数数列an,的前n项和为Sn,且对任意正整数n满足Sn=1/4(an+1)^2。

Answer 1

解：
1、
a1=S1=(1/4)(a1+1)^2
4a1=a1^2+2a1+1
(a1-1)^2=0
a1=1
Sn=(1/4)(an+1)^2
Sn-1=(1/4)[a(n-1)+1]^2
an=Sn-Sn-1=(1/4)[an^2+2an+1-a(n-1)^2-2a(n-1)-1]
4an=an^2+2an-a(n-1)^2-2a(n-1)
an^2-2an=a(n-1)^2+2a(n-1)
(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为正数数列，舍去)
{an}是首项为1，公差为2的等差数列。
an=1+2(n-1)=2n-1
{an}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)]
Bn=(1/2)[1/1-1/3+1/3-1/5+...+1/(2n-3)-1/(2n-1)+1/(2n-1)-1/(2n+1)]
=(1/2)[1-1/(2n+1)]
=n/(2n+1)