Question

求解一不等式`急!已知数列{an}的前n项和有Sn,首项为a1,且1,an,sn成等差数列（1）求数列{an}的通项公式

Answer 1

首项为a1,且1,an,Sn成等差数列

2an=1+sn

2a(n-1)=1+s(n-1)

相减得

2[an-a(n-1)]=an

an/a（n-1）=2

an=2^(n-1)

2)设Tn为数列1/an的前n项和

Tn=1/2^0+2/2^1+3/2^2+...+n/2^(n-1)

Tn/2=1/2^1+2/2^2+3/2^3+...+n/2^n

Tn-Tn/2

=1/2^0+1/2^1+1/2^2+...+1/2^(n-1)-n/2^n

=[1-(1/2)^n]/(1-1/2)-n/2^n

=2*[1-(1/2)^n]-n/2^n

Tn=4*[1-(1/2)^n]-2n/2^n

=4-4/2^n-2n/2^n

=4-(4+2n)/2^n

1==3

m>=7

m的最小值:7