高一数学数列问题求解

(1)
an=2a(n-1)+2^n -1
an/2^n - a(n-1)/2^(n-1) = 1 - 1/2^n
an/2^n-a4/2^4 = (n-4) - (1/2^5+1/2^6+...+1/2^n)
an/2^n -81/16 = (n-4) - (1/2^4)[ 1- (1/2)^(n-4) ]
= n- 17/16 + (1/2)^n
an = (n+4).2^n + 1
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)

Sn = a1+a2 +...+an
= S + 8(2^n - 1) + n
=n.2^(n+1) +6(2^n -1) +n
= n-6 +(2n+6).2^n
(2)
bn = (an+ λ)/ 2^n
λ =-1
bn = n+4
=>{bn}是等差数列