Question

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

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)