已知数列an中,a1=1,a(n+1)=an/(an+3)(n属于N*)数列bn满足bn=（3^n-1)n/2^n*an,求数列bn的前n项和

Question

已知数列an中,a1=1,a(n+1)=an/(an+3)(n属于N*)
数列bn满足bn=（3^n-1)n/2^n*an,求数列bn的前n项和

Answer 1

a(n+1)=an/(an+3)
a(n+1)+2 =an/(an+3)+2
= 3(an+2)/(an+3)
1/[a(n+1)+2] = (an+3)[3(an+2)]
= 1/3 + (1/3)[ 1/(an+2)]
1/[a(n+1)+2] -1/2 = (1/3) { [ 1/(an+2)] - 1/2 }
=>{ [ 1/(an+2)] - 1/2 }是等比数列,q=1/3
{ [ 1/(an+2)] - 1/2 }= (1/3)^(n-1).{ [ 1/(a1+2)] - 1/2 }
=-(1/2).(1/3)^n
1/(an+2) =(1/2)( 1- (1/3)^n)
an +2 = 2/( 1- (1/3)^n)
an = -2 +2/( 1- (1/3)^n)
= -2 + 2.3^n/(3^n -1)
= 2/(3^n -1)
let
S = 1.(1/2)^1+2.(1/2)^2+...+n.(1/2)^n (1)
(1/2)S = 1.(1/2)^2+2.(1/2)^3+...+n.(1/2)^(n+1) (2)
(1)-(2)
(1/2)S = (1/2+1/2^2+...+1/2^n) -n.(1/2)^(n+1)
= (1- (1/2)^n ) - n.(1/2)^(n+1)
bn=(3^n-1)n/(2^n*an)
=n/2^(n+1)
= (1/2)[ n.(1/2)^n ]
Tn =b1+b2+...+bn
=(1/2)S
= (1- (1/2)^n ) - n.(1/2)^(n+1)
= 1- (n+2)(1/2)^(n+1)