求证等差数列,a1=1 ,an=2a(n-1)+ 2^(n-1) 设bn= an/2^(n-1) 求证bn是等差数列

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求证等差数列,a1=1 ,an=2a(n-1)+ 2^(n-1) 设bn= an/2^(n-1) 求证bn是等差数列

求证等差数列,a1=1 ,an=2a(n-1)+ 2^(n-1) 设bn= an/2^(n-1) 求证bn是等差数列
求证等差数列,
a1=1 ,an=2a(n-1)+ 2^(n-1) 设bn= an/2^(n-1) 求证bn是等差数列

求证等差数列,a1=1 ,an=2a(n-1)+ 2^(n-1) 设bn= an/2^(n-1) 求证bn是等差数列
bn=an/2^(n-1)
bn-1=an-1/2^(n-2)
bn-bn-1= an/2^(n-1)-an-1/2^(n-2)
=an/2^(n-1)-2an-1/2^(n-1)
=an-2an-1/2^(n-1)
=2an-1+2^(n-1)-2an-1/2^(n-1)
=1
∴bn是等差数列

bn= an/2^(n-1) =2a(n-1)/2^(n-1)+1=2x2a(n-2)/2^(n-1)+1+1=2a(n-2)/2^(n-2)+1+1
bn-1= an-1/2^(n-2) =2a(n-2)/2^(n-2)+1
bn-bn-1=1

等式两边同除以2^(n-1)得:an/(2^(n-1))=a(n-1)/(2^(n-2))+1,即bn=b(n-1)+1(n>=2),所以从第一项开始成等差,d=1

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