数列 an 的前n项和为sn,若向量a=(Sn,1),b=(-1,2an+2^n+1),a⊥b,{bn}=an/2^n

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数列 an 的前n项和为sn,若向量a=(Sn,1),b=(-1,2an+2^n+1),a⊥b,{bn}=an/2^n
(1)求证数列是{bn}等差数列,并求出数列{an}的通项公式
（2）求数列{an}的前n项和Sn

(1)
a=(Sn,1),b=(-1,2an+2^(n+1))
a.b=0
-Sn+2an + 2^(n+1) =0
Sn =2an + 2^(n+1)
n=1
a1= -4
an = Sn -S(n-1)
= 2an - 2a(n-1) + 2^n
an = 2a(n-1) -2^n
an/2^n-a(n-1)/2^(n-1) = -1
an/2^n-a1/1=-(n-1)
an/2^n = -(n+3)
bn=an/2^n = -(n+3)
an = -(n+3).2^n
(2)
Sn =2an + 2^(n+1)
=-2(n+3).2^n +2^(n+1)
=-2(n+2).2^n
再问： an/2^n-a(n-1)/2^(n-1) = -1
an/2^n-a1/1=-(n-1)
这两步是怎么划出来的复制去Google翻译翻译结果
再答：不好意思，错了
an/2^n-a(n-1)/2^(n-1) = -1
=>{an/2^n}是等差数列,d=-1

an/2^n-a1/2=-(n-1)
an/2^n = -(n+1)
bn=an/2^n = -(n+1)
an = -(n+1).2^n
bn=an/2^n = -(n+1)
an = -(n+1).2^n
(2)
Sn =2an + 2^(n+1)
=-2(n+1).2^n +2^(n+1)
=-n.2^(n+1)

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