设数列an前n项和为Sn,且Sn=2^n -1 数列bn满足b1=2,b（n+1)-2bn=8an

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设数列an前n项和为Sn,且Sn=2^n -1 数列bn满足b1=2,b（n+1)-2bn=8an
证明：数列{bn/2^2}为等差数列,并求bn的通项公式及前n项和Tn
本人才疏学浅,麻烦过程清楚一些

n=1时,a1=S1=2^1 -1=2-1=1
n≥2时,
Sn=2^n -1
Sn-1=2^(n-1) -1
Sn-Sn-1=an=2^n -1-2^(n-1) +1=2^(n-1)
n=1时,a1=2^(1-1)=2^0=1,同样满足.
数列{an}的通项公式为an=2^(n-1)
b(n+1)-2bn=8×2^(n-1)=2^(n+2)
b(n+1)=2bn+2^(n+2)
b(n+1)/2^(n+1)=2bn/2^(n+1)+2^(n+2)/2^(n+1)
b(n+1)/2^(n+1)=bn/2^n +2
[b(n+1)/2^(n+1)]-(bn/2^n)=2,为定值.
b1/2^1=2/2=1
数列{bn/2^n}是以1为首项,2为公差的等差数列.
bn/2^n=1+2(n-1)=2n-1
bn=(2n-1)×2^n
n=1时,b1=(2-1)×2^1=2,同样满足.
数列{bn}的通项公式为bn=(2n-1)×2^n.
Tn=b1+b2+...+bn=1×2^1+3×2^2+5×2^3+...+(2n-1)×2^n
2Tn=1×2^2+3×2^3+...+(2n-3)×2^n+(2n-1)×2^(n+1)
Tn-2Tn=-Tn=2^1+2×2^2+2×2^3+...+2×2^n-(2n-1)×2^(n+1)
=2+2×4×[2^(n-1) -1]/(2-1) -(2n-1)×2^(n+1)
=(3-2n)×2^(n+1)-6
Tn=(2n-3)×2^(n+1) +6.
^表示指数.

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