当 n = 1时,Sn = a1.则 a1 = 2/3因为 Sn + 1/2*an = 1,所以有：(Sn + 1/2*an) - [S(n-1) + 1/2*a(n-1)] = 0[Sn - S(n-1)] + 1/2*[an - a(n-1)] = 0an + 1/2*an - 1/2*[a(n-1)] = 0an/[a(n-1)] = 1/3即 an 是一个等比数列.它的通项式：an = a1 * (1/3)^(n-1) = 2*(1/3)^n所以,Sn = 1 - (1/3)^n = 1 - 3^(-n)bn = - log3 [3^(-n)] = n所以：Tn = (1/b1-1/b2) + (1/b2 - 1/b3) + …+ (1/bn - 1/b(n+1)) = 1 - 1/(n+1) = n/(n+1)
