设数列{an}的前n项和为Sn且a1=1,Sn+1=4an+2(n属于正整数) (1)设bn=an/2n,求证数列{bn}是等差数列 (2设数列{an}的前n项和为Sn且a1=1,Sn+1=4an+2(n属于正整数)(1)设bn=an/2n,求证数列{bn}是等差数列(2)求
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![设数列{an}的前n项和为Sn且a1=1,Sn+1=4an+2(n属于正整数) (1)设bn=an/2n,求证数列{bn}是等差数列 (2设数列{an}的前n项和为Sn且a1=1,Sn+1=4an+2(n属于正整数)(1)设bn=an/2n,求证数列{bn}是等差数列(2)求](/uploads/image/z/13907265-33-5.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%E4%B8%94a1%3D1%2CSn%2B1%3D4an%2B2%28n%E5%B1%9E%E4%BA%8E%E6%AD%A3%E6%95%B4%E6%95%B0%29+%EF%BC%881%EF%BC%89%E8%AE%BEbn%3Dan%2F2n%2C%E6%B1%82%E8%AF%81%E6%95%B0%E5%88%97%7Bbn%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97+%EF%BC%882%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%E4%B8%94a1%3D1%2CSn%2B1%3D4an%2B2%28n%E5%B1%9E%E4%BA%8E%E6%AD%A3%E6%95%B4%E6%95%B0%29%EF%BC%881%EF%BC%89%E8%AE%BEbn%3Dan%2F2n%2C%E6%B1%82%E8%AF%81%E6%95%B0%E5%88%97%7Bbn%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BC%882%EF%BC%89%E6%B1%82)
设数列{an}的前n项和为Sn且a1=1,Sn+1=4an+2(n属于正整数) (1)设bn=an/2n,求证数列{bn}是等差数列 (2设数列{an}的前n项和为Sn且a1=1,Sn+1=4an+2(n属于正整数)(1)设bn=an/2n,求证数列{bn}是等差数列(2)求
设数列{an}的前n项和为Sn且a1=1,Sn+1=4an+2(n属于正整数) (1)设bn=an/2n,求证数列{bn}是等差数列 (2
设数列{an}的前n项和为Sn且a1=1,Sn+1=4an+2(n属于正整数)
(1)设bn=an/2n,求证数列{bn}是等差数列
(2)求数列{an}的通项公式及前n项和的公式
设数列{an}的前n项和为Sn且a1=1,Sn+1=4an+2(n属于正整数) (1)设bn=an/2n,求证数列{bn}是等差数列 (2设数列{an}的前n项和为Sn且a1=1,Sn+1=4an+2(n属于正整数)(1)设bn=an/2n,求证数列{bn}是等差数列(2)求
因为S(n+1)=4an+2 一式
n>=2时,Sn=4a(n-1)+2 二式
所以一式减二式,得 a(n+1)=4an-4a(n-1)
(目标是a(n+1)+m*an=K(an+m*a(n-1)),所以构建等比数列如下)
a(n+1)=(K-m)an=K(an+m*a(n-1))
可得K-m=4 m*k=-4
所以K=2,m=-2
所以a(n+1)-2an=2*(an-2a(n-1))
所以an-2*a(n-1)是一个以2为公比的等比数列
则an-2*a(n-1)=(a2-2*a1)*2^(n-2)
当n=1时,a1+a2=4(a1)+2
所以a2=5
an-2*a(n-1)=3*2^(n-2) 同除以2
得an/2^n-2*a(n-1)/2^n=(3*2^(n-2))/2^n=3/4
设bn=an比2的n次方
即bn-b(n-1)=3/4
所以数列bn是一个以3/4为公差的等差数列
即得证
(1)Sn+1=4an+2
Sn=4a(n-1)+2
相减得Sn+1-Sn=4an+2-4a(n-1)-2
an+1=4an-4a(n-1)
an+1-2an=2(an-2an-1)
bn=2bn-1