n = 1時(shí)
S1 + a1 = 2 * a1 = 2,所以a1 = 1.
將n替換成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)
這個(gè)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))
把這里面的n+1再換成n就是第一問的式子
（2）由第一問,{a(n) - (2n-1)}構(gòu)成一個(gè)公比為1/2的等比數(shù)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))（裂項(xiàng)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)