已知数列{an}的前n项和为Sn,且Sn+an=n^2+2n-1对于所有自然数n恒成立1、求证：an-(2n-1)=[a(n-1)-(2n-3)]/2,求数列{an}的通项公式2、若数列{bn}满足：bn=1/a(n+1)an,求b1+b2+…+bn的和

啊！！！这么难，另请高人吧

无解

n = 1时
S1 + a1 = 2 * a1 = 2,所以a1 = 1.
将n替换成n+1,注意到Sn+1 = Sn + a(n+1)
所以Sn+1 + a(n+1) = (n+1)^2 + 2(n+1) - 1
= Sn + 2*a(n+1) = (n^2 + 2n - 1 - an ) + 2 * a(n+1)
这个式子整理一下,得2n+3 + a(n) = 2 * a(n+1)
所以a(n) - (2n-1) = 2( a(n+1) - (2n+1))
把这里面的n+1再换成n就是第一问的式子
（2）由第一问,{a(n) - (2n-1)}构成一个公比为1/2的等比数列,所以
a(n) - (2n-1) = (1/2)^(n-1) * (a1 - (2*1-1)) = 0
所以a(n) = 2n-1
bn = 1/(2n-1)(2n+1) = 1/2 * (1/(2n-1) - 1/(2n+1))（裂项求和）
所以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)