求下题详解谢谢

a1=1 , a(n+1)= an/(an+3)
(1)
a(n+1)= an/(an+3)
n=1, a2= a1/(a1+3) = 1/(1+3) = 1/4
n=2, a3= a2/(a2+3) = (1/4)/(1/4+3) = 1/7
(2)
a(n+1)= an/(an+3)
1/a(n+1) = (an+3)/an
= 1 + 3/an
1/a(n+1) + 1/2 = 3 ( 1/an + 1/2 )
=> { 1/an + 1/2 } 是等比数列, q=3
1/an + 1/2 = 3^(n-1) . ( 1/a1 + 1/2 )
= (1/2) . 3^n
1/an = -1/2 +(1/2) . 3^n
an = 2/(-1 + 3^n )
(3)
bn
= (3^n -1). (n/2^n).an
= (3^n -1). (n/2^n). [ 2/(-1 + 3^n ) ]
=n/2^(n-1)
let
S = 1.(1/2)^0 +2.(1/2)^1+...+n.(1/2)^(n-1) (1)
(1/2)S = 1.(1/2)^1 +2.(1/2)^2+...+n.(1/2)^n (2)
(1)-(2)
(1/2)S = [ 1/2^0+1/2^1+...+1/2^(n-1)] -n.(1/2)^n
= 2( 1- 1/2^n) - n.(1/2)^n
S = 4( 1- 1/2^n) - 2n.(1/2)^n
Tn
=b1+b2+...+bn
=S
= 4( 1- 1/2^n) - 2n.(1/2)^n
=4 - (2n+4).(1/2)^n

(1) a2 = a1/a1+3 = 1/4, a3 = a2/a2+3 = 1/13
(2)由已知条件，1/an+1 = an+3/an = 1+3/an
1/an+1 + 1/2 = 3/2 + 3/an = 3(1/an +1/2)
因此{1/an + 1/2}是公比为3的等比数列
{1/an + 1/2} = 3/2* 3^(n-1)
计算an = 1/(3^n/2 -1/2) =2/(3^n-1)
(3)由(2)结果，带入bn = n/2^n-1
Tn=1/2^0 + 2/2^1 ... n/2^n-1
1/2 Tn = 1/2^1 + 2/2^2 + ... + n/2^n
两式相减

1/2 Tn = 1/2^0 +[1/2^1 +1/2^2...+1/2^n-1] - n/2^n
= 1/2^0 + (1-1/2^n-1) - n/2^n
Tn = 4-(2+n)/2^n-1