(1)
a(n+1)=2an/(an +1)
1/a(n+1)=(an+1)/(2an)=(1/2)(1/an)+1/2
1/a(n+1) -1=(1/2)(1/an)-1/2=(1/2)(1/an -1)
[1/a(n+1)-1]/(1/an -1)=1/2,为定值.
1/a1-1=1/(2/3)-1=1/2
数列{1/an -1}是以1/2为首项,1/2为公比的等比数列.
(2)
1/an -1=1/2ⁿ
1/an=1+1/2ⁿ
(2n-1)/an=(2n-1)(1+1/2ⁿ)=(2n-1)+(2n-1)/2ⁿ=2n-1-1/2ⁿ+n/2^(n-1)
Sn=2(1+2+...+n)-n-(1/2)(1-1/2ⁿ)/(1-1/2)+1/1+2/2+3/2²+...+n/2^(n-1)
=n²+1/2ⁿ-1+[1+2/2+3/2²+...+n/2^(n-1)]
令Cn=1+2/2+3/2²+...+n/2^(n-1)
则Cn/2=1/2+2/2²+...+(n-1)/2^(n-1)+n/2ⁿ
Cn-Cn/2=Cn/2=1+1/2+1/2²+...+1/2^(n-1)-n/2ⁿ
=(1-1/2ⁿ)/(1-1/2)-n/2ⁿ
=2(1-1/2ⁿ)-n/2ⁿ
Cn=4-4/2ⁿ-2n/2ⁿ
Sn=n²+1/2ⁿ-1+4-4/2ⁿ-2n/2ⁿ=n²-(2n+3)/2ⁿ +3