Question

已知数列{an}中，a1=1,sn是{an}的前n项和，当n≥2sn=an(1-sn/2)，

①求证{sn/1}是等差数列 ②若Tn=S1·S2+S2·S3+…+Sn·Sn+1,Tn在条件②下，试求满足不等式am+1+am+2+…+a2m/2m≥-2/77T5的正整数m

Answer 1

① Sn=an(1-2/Sn)
因an=Sn-S(n-1)
所以Sn=[Sn-S(n-1)]*(1-2/Sn)=Sn-S(n-1)-2+2S(n-1)/Sn
两边同除以2S(n-1)
1/Sn-1/S(n-1)=1/2
所以{1/Sn}是公差为1/2的等差数列
② 1/Sn=1/S1+(1/2)(n-1)=(n+1)/2
所以Sn=2/(n+1)
Sn*S(n+1)=4/[(n+1)(n+2)]=4[1/(n+1)-1/(n+2)]
所以Tn=4{(1/2-1/3)+(1/3-1/4)+......+[1/(n+1)-1/(n+2)]}
=4*[1/2-1/(n+2)]
=2n/(n+2)
an=Sn-S(n-1)=2/(n+1)-2/n=2[1/(n+1)-1/n]
[a(m+1)+a(m+2)+...+a2m]/2m
=2[1/(m+2)-1/(m+1)+1/(m+3)-1/(m+2)+.....+1/(2m+1)-1/(2m)]
=2[1/(m+2)-1/(2m)]
≥-2/[77*2*5/(5+2)]
=-1/55
解得m≥2
所以最小的m=2满足条件

Answer 2

{sn/1}？{1/sn}？