设 数列{an}的前n项和为Sn,已知b*an - 2^n=(b-1)Sn求证:当b=2时,{an-n*2^(n-1) } 是等比数列解析:由题意得,a1=2,且b*an-2^n=(b-1)Sn,b*an+1 - 2^(n+1) =(b - 1) Sn+1,两式相减得b(an+1 - n*2^(n-1) ) - 2^n = (b-1)*an+ 1主
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![设 数列{an}的前n项和为Sn,已知b*an - 2^n=(b-1)Sn求证:当b=2时,{an-n*2^(n-1) } 是等比数列解析:由题意得,a1=2,且b*an-2^n=(b-1)Sn,b*an+1 - 2^(n+1) =(b - 1) Sn+1,两式相减得b(an+1 - n*2^(n-1) ) - 2^n = (b-1)*an+ 1主](/uploads/image/z/15209156-20-6.jpg?t=%E8%AE%BE+%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E5%B7%B2%E7%9F%A5b%2Aan+-+2%5En%3D%28b-1%29Sn%E6%B1%82%E8%AF%81%EF%BC%9A%E5%BD%93b%3D2%E6%97%B6%2C%7Ban-n%2A2%5E%28n-1%29+%7D+%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%E8%A7%A3%E6%9E%90%EF%BC%9A%E7%94%B1%E9%A2%98%E6%84%8F%E5%BE%97%2Ca1%3D2%2C%E4%B8%94b%2Aan-2%5En%3D%28b-1%29Sn%2Cb%2Aan%2B1+-+2%5E%28n%2B1%29+%3D%28b+-+1%29+Sn%2B1%2C%E4%B8%A4%E5%BC%8F%E7%9B%B8%E5%87%8F%E5%BE%97b%28an%2B1+-+n%2A2%5E%28n-1%29+%EF%BC%89+-+2%5En+%3D+%28b-1%29%2Aan%2B+1%E4%B8%BB)
设 数列{an}的前n项和为Sn,已知b*an - 2^n=(b-1)Sn求证:当b=2时,{an-n*2^(n-1) } 是等比数列解析:由题意得,a1=2,且b*an-2^n=(b-1)Sn,b*an+1 - 2^(n+1) =(b - 1) Sn+1,两式相减得b(an+1 - n*2^(n-1) ) - 2^n = (b-1)*an+ 1主
设 数列{an}的前n项和为Sn,已知b*an - 2^n=(b-1)Sn
求证:当b=2时,{an-n*2^(n-1) } 是等比数列
解析:由题意得,a1=2,且b*an-2^n=(b-1)Sn,
b*an+1 - 2^(n+1) =(b - 1) Sn+1,两式相减得
b(an+1 - n*2^(n-1) ) - 2^n = (b-1)*an+ 1
主要是这边不知道:为什么 b*an-2^n=(b-1)Sn 减去 b*an+1 - 2^(n+1) =(b - 1) Sn+1
会等于b(an+1 - n*2^(n-1) ) - 2^n = (b-1)*an+ 1 主要是不知道为什么那边的
2^(n+1) - 2^n 会等于 -2^n
设 数列{an}的前n项和为Sn,已知b*an - 2^n=(b-1)Sn求证:当b=2时,{an-n*2^(n-1) } 是等比数列解析:由题意得,a1=2,且b*an-2^n=(b-1)Sn,b*an+1 - 2^(n+1) =(b - 1) Sn+1,两式相减得b(an+1 - n*2^(n-1) ) - 2^n = (b-1)*an+ 1主
2^(n+1) - 2^n
=2*2^n - 2^n
=2^n
b*an-2^n=(b-1)Sn,
b*a(n+1)- 2^(n+1)=(b-1)S(n+1)
两式相减(左-左=右-右):
[b*a(n+1)- 2^(n+1)]-[b*an-2^n]=[(b-1)S(n+1)]-[(b-1)Sn]
[b*a(n+1)-b*an]-[2^(n+1)-2^n]=(b-1)[S(n+1)-Sn]
ba(n+1)-ban-2^n=(b-1)a(n+1)
a(n+1)-ban-2^n=0
我觉得相减得到的b(an+1 - n*2^(n-1) ) - 2^n = (b-1)*an+ 1似乎算得不对