设正数数列an前n项和为Sn,Sn=二分之一(an+an分之一),求an的通项公式

题目

设正数数列an前n项和为Sn,Sn=二分之一(an+an分之一),求an的通项公式

答案

Sn = (1/2)(an + 1/an) (1)
n=1
2(a1)^2 = (a1)^2+a1
a1(a1-1)=0
a1 =1

S(n-1) = (1/2)[a(n-1) +1/a(n-1) ] (2)
(1)-(2)
an = (1/2)(an + 1/an)- (1/2)[a(n-1) +1/a(n-1) ]
1/an - an - a(n-1) -1/a(n-1) =0
a(n-1) - 2an.a(n-1) - an =0
1/an -1/a(n-1) =2
1/an -1/a1 = 2(n-1)
1/an = 2n-1
an = 1/(2n-1)