Question

求好心人！

Answer 1

a1=1 ,a(n+1)＝1-（1/4an） ,bn=2/[2an-1]
a(n+1)＝(4an-1)/4an
2a(n+1)＝(4an-1)/2an
2a(n+1)-1＝[(4an-1)/2an]-1
2a(n+1)-1＝[(2an-1)/2an]
倒数得
1/[2a(n+1)-1]＝2an/(2an-1)
1/[2a(n+1)-1]＝1+[1/(2an-1)]
1/[2a(n+1)-1]-[1/(2an-1)]=1
{1/(2an-1)}是等差数列
1/(2an-1)＝1/(2a1-1)+n-1=1/(2-1)+n-1=n
1/(2an-1)=n
an=(n+1）/2n
bn-b(n-1)＝2/(2an-1)-2/[2a(n-1)-1]=2[1/(2an-1)-1/[2a(n-1)-1]
=2{1/[(n+1）/n-1]-1/1/[n/(n-1)-1]}
=2[n-(n-1）]
＝2
所以，bn-b(n-1)＝2
{bn}是等差数列
2)Cn=4an/(n+1)=4(n+1）/2n(n+1)=2/n
Cn=2/n
CnC(n+2)=2/n*2/(n+2）＝4/n(n+2)=2[1/n-1/(n+2)]

Tn=2[(1-1/3)+(1/2-1/4)+(1/3-1/5)+(1/4-1/6)+（1/5-1/7）+...+1/n-1/(n+2)]
Tn=2[1+1/2-1/(n+1)-1/(n+2）]=2[3/2-(2n+3)/(n^2+3n+2）
Tn=3-[(2n+3)/2(n^2+3n+2）]<3
所以Tn<3