已知数列｛an｝满足:a1=1/2，1/2an+1(小一)=1/2an +大1(n属于N*)

已知数列｛an｝满足:a1=1/2，1/2an+1(小一)=1/2an +大1(n属于N*) (1)求证:数列｛1/an｝是等差数列;(2)设bn=2^n/an，Tn为数列｛bn｝的前n项和，求Tn

(1)
1/(2a(n+1))=1/(2an) +1
1/a(n+1) = 1/an + 2
1/a(n+1)-1/an = 2
{1/an}是等差数列, d=2
(2)
1/an -1/a1= 2(n-1)
1/an =2n
an = 1/(2n)
bn=2^n/an
= 2(n.2^n)
let
S=1.2^1+2.2^2+....+n.2^n (1)
2S= 1.2^2+2.2^3+....+n.2^(n+1) (2)
(2)-(1)
S = n.2^(n+1) -(2+2^2+...+2^n)
=n.2^(n+1) - 2(2^n-1)
= 2 - (2n-2).2^n
Tn =b1+b2+...+bn
= 2S
= 4- (4n-4).2^n

因为： a(n+1) + 1 = (2an + 1) + 1 = 2(an + 1) [a(n+1) + 1]/(an + 1) = 2 1. 可以看出，新的数列{an + 1} 是一个等比数列，公比 q = 2 2. an + 1 = (a1 + 1) * q^(n-1) = 2 * 2^(n-1) = 2^n 所以，an = 2^n - 1 3. sn = (2^1 - 1) + (2^2 - 1) + ……+ (2^n - 1)      = (2^1 + 2^2 + ……+ 2^n) - n      = [2^(n+1) - 2^1]/(2 - 1) - n      = 2^(n+1) - (n + 2)