已知数列{an}和{bn}满足a1=m,an+1=λan+n,bn=an-2n/3+4/9已知数列{an}和{bn}满足a1=m,a(n+1)=λan+n,bn=an-2n/3+4/9(1)当m=1时,求证:于任意的实数λ,{an}一定不是等差数列(2)当λ=-1/2时,试判断{bn}

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已知数列{an}和{bn}满足a1=m,an+1=λan+n,bn=an-2n/3+4/9已知数列{an}和{bn}满足a1=m,a(n+1)=λan+n,bn=an-2n/3+4/9(1)当m=1时

已知数列{an}和{bn}满足a1=m,an+1=λan+n,bn=an-2n/3+4/9已知数列{an}和{bn}满足a1=m,a(n+1)=λan+n,bn=an-2n/3+4/9(1)当m=1时,求证:于任意的实数λ,{an}一定不是等差数列(2)当λ=-1/2时,试判断{bn}
已知数列{an}和{bn}满足a1=m,an+1=λan+n,bn=an-2n/3+4/9
已知数列{an}和{bn}满足a1=m,a(n+1)=λan+n,bn=an-2n/3+4/9
(1)当m=1时,求证:于任意的实数λ,{an}一定不是等差数列
(2)当λ=-1/2时,试判断{bn}是否为等比数列

已知数列{an}和{bn}满足a1=m,an+1=λan+n,bn=an-2n/3+4/9已知数列{an}和{bn}满足a1=m,a(n+1)=λan+n,bn=an-2n/3+4/9(1)当m=1时,求证:于任意的实数λ,{an}一定不是等差数列(2)当λ=-1/2时,试判断{bn}
(1)分情况讨论
当λ=0时求出b1=1,b2=1,b3=2从这三项就可看出数列{an}不是等差数列
当λ≠0时这种情形用数学归纳法
a3+a1-2a2=λa2+2+1-2(λ+1)
=λ*(λ+1)+3-2(λ+1)
=λ^2-λ+1
=(λ-1/2)^2+3/4≠0
即a3+a1≠2a2
a4+a2-2a3=(λa3+3)+a2-2a3
=(λ-2)a3+a2+3
=(λ-2)(λa2+2)+a2+3
=(λ+1)(λ^2-2λ+1)+2(λ-2)+3
=λ(λ^2-λ+1)
=λ[(λ-1/2)^2+3/4]≠0
即a4+a2≠2a3
假设a(k+2)+ak≠2a(k+1)成立
那么a(k+3)+a(k+1)-2a(k+2)=λa(k+2)+k+2+λak+k-2[λa(k+1)+k+1]
=λ[a(k+2)+ak-2a(k+1)]≠0
所以当λ≠0时数列{an}不是等差数列
综上所述当m=1时对于任意的实数λ,{an}一定不是等差数列
(2)当λ=-1/2时则a(n+1)=-)=-an/2+n
b(n+1)/bn={a(n+1)-[2(n+1)]/3+4/9}/(an-2n/3+4/9)
=(-an/2+n-2n/3-2/9)/(an-2n/3+4/9)
=(-1/2)(an-2n/3+4/9)/(an-2n/3+4/9)
=-1/2
所以数列{bn}是以b1=m-2/9为首项,以-1/2为公比的等比数列

(2)是

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