Question

已知函数f(x)=x/(3x+1),数列{An}满足A1=1,A(n+1)=f(An)

已知函数f(x)=x/(3x+1),数列{An}满足A1=1,A(n+1)=f(An)(n属于N*).数列{Bn}的前n项和Sn=2^n-1.(1)求证：数列{1/An}为等差数列；(2)设数列Cn=Bn/An,求数列{Cn}的前n项和Tn;(3)若1/(An*Bn)小于等于m^2-2m-1,m属于N*恒成立，求m的最小值。

Answer 1

(1)
a(n+1)= f(an)
= an/(3an+1)
1/a(n+1) = 3+1/an
1/a(n+1) -1/an=3
1/an -1/a1=3(n-1)
=>[1/an}}为等差数列

(2)
1/an = 3n-2
an = 1/(3n-2)
Sn=b1+b2+..+bn = 2^n-1
bn =sn-S(n-1) = 2^(n-1)
cn =bn/an
= (3n-2)2^(n-1)
= 3(n.2^(n-1)) - 2^n
Tn =c1+c2+..+cn
= 3{summation(i:1->n) i.2^(i-1) } - 2(2^n-1)
consider
1+x+x^2+..+x^n = (x^(n+1)- 1)/(x-1)
1+2x+..+nx^(n-1) =[(x^(n+1)- 1)/(x-1)]'
= [nx^(n+1) - (n+1)x^n + 1]/(x-1)^2
put x=2
{summation(i:1->n) i.2^(i-1) }
=[n.2^(n+1) - (n+1).2^n + 1]
= 1+ (n-1)2^n

Tn =3{summation(i:1->n) i.2^(i-1) } - 2(2^n-1)
=3[1+ (n-1)2^n ] - 2(2^n-1)
=5+ (3n-5).2^n
(3)
1/(an*bn)<=m^2-2m-1 ,m属于N*恒成立
m属于N*恒成立, 求m的最小值 ?????