Question

设等比数列{an}的前n项和为Sn满足:a(n+1)=a1Sn+1

Ⅰ求a1的值及数列｛an｝的通项公式
Ⅱ设bn=2^n/{(an+1)(a(n+1)+1)},数列{bn}的前n项和Tn，且满足Tn≤Y恒成立，求实数Y的最小值

Answer 1

a2=a1*a1+1=(a1)^2+1;
a3=a1(a1+a2)+1=(a1)^2+a1a2+1=a2+a1a2=a2(a1+1)
(a2)^2=a1*a3=a1*a2(a1+1)
a2=a1(a1+1) (a1)^2+1=a1(a1+1) a1=1

a2=1+1=2
a3=2*2=4
an=2^(n-1)
bn=2^n/{(2^(n-1)+1)(2^n+1)}
bn=2(1/(2^(n-1)+1)-1/(2^n+1)}
Tn=2[1/2-1/3+1/3-1/5+....+1/(2^(n-2)+1)-1/(2^(n-1)+1)+1/(2^(n-1)+1)-1/(2^n+1)]
Tn=2[1/2-1/(2^n+1)]<1
Y最小值为1