已知数列an满足a1=1,an-2an-1-2^(n-1)=0(n属于正整数，n≥2，设bn=an/2^n

Question

求b1,b2的值，并证明数列bn是等差数列
求数列an的前n项和Sn

Answer 1

an -2a(n-1) -2^(n-1)=0
an/2^n - a(n-1)/2^(n-1) = 1/2
bn=an/2^n 是等差数列, d=1/2
an/2^n - a1/2^1 = (n-1)/2
an/2^n = n/2
an = n.2^(n-1)

b1= 1/2
b2= 1
let
S = 1.2^0+2.2^1+....+n.2^(n-1) (1)
2S = 1.2^1+2.2^2+....+n.2^n (2)
(2)-(1)
S = n.2^n - [1+2+...+2^(n-1) ]
=n.2^n - (2^n-1)
= 1 + (n-1).2^n
Sn = a1+a2+...+an
= S
= 1 + (n-1).2^n