作业帮已知数列{an}是首项为a1>0,公比q>-1的等比数列,若数列{bn}通项bn=a[n+1]-ka[n+2](n∈N+）已知数列{an}是首项为a1>0,公比q>-1的等比数列,若数列{bn}通项bn=a[n+1]-ka[n+2](n∈N+）,数列{an}{bn}的前n项和分别为Sn
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![已知数列{an}是首项为a1>0,公比q>-1的等比数列,若数列{bn}通项bn=a[n+1]-ka[n+2](n∈N+）已知数列{an}是首项为a1>0,公比q>-1的等比数列,若数列{bn}通项bn=a[n+1]-ka[n+2](n∈N+）,数列{an}{bn}的前n项和分别为Sn](/uploads/image/z/5311725-69-5.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%98%AF%E9%A6%96%E9%A1%B9%E4%B8%BAa1%3E0%2C%E5%85%AC%E6%AF%94q%3E-1%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E8%8B%A5%E6%95%B0%E5%88%97%7Bbn%7D%E9%80%9A%E9%A1%B9bn%3Da%5Bn%2B1%5D-ka%5Bn%2B2%5D%28n%E2%88%88N%2B%EF%BC%89%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%98%AF%E9%A6%96%E9%A1%B9%E4%B8%BAa1%3E0%2C%E5%85%AC%E6%AF%94q%3E-1%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E8%8B%A5%E6%95%B0%E5%88%97%7Bbn%7D%E9%80%9A%E9%A1%B9bn%3Da%5Bn%2B1%5D-ka%5Bn%2B2%5D%28n%E2%88%88N%2B%EF%BC%89%2C%E6%95%B0%E5%88%97%7Ban%7D%7Bbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E5%88%86%E5%88%AB%E4%B8%BASn)
已知数列{an}是首项为a1>0,公比q>-1的等比数列,若数列{bn}通项bn=a[n+1]-ka[n+2](n∈N+）已知数列{an}是首项为a1>0,公比q>-1的等比数列,若数列{bn}通项bn=a[n+1]-ka[n+2](n∈N+）,数列{an}{bn}的前n项和分别为Sn
已知数列{an}是首项为a1>0,公比q>-1的等比数列,若数列{bn}通项bn=a[n+1]-ka[n+2](n∈N+）
已知数列{an}是首项为a1>0,公比q>-1的等比数列,若数列{bn}通项bn=a[n+1]-ka[n+2](n∈N+）,数列{an}{bn}的前n项和分别为Sn和Tn,如果Tn>kSn对一切正整数n都成立,求实数k的取值范围.
a[n+1]-ka[n+2]中[n+1]、[n+2]都是下脚标
题中说了q>-1!
已知数列{an}是首项为a1>0,公比q>-1的等比数列,若数列{bn}通项bn=a[n+1]-ka[n+2](n∈N+）已知数列{an}是首项为a1>0,公比q>-1的等比数列,若数列{bn}通项bn=a[n+1]-ka[n+2](n∈N+）,数列{an}{bn}的前n项和分别为Sn
楼上几位的分类不完整额.
an=a1q^(n-1)
则Sn=a1(1-q^n)/(1-q),由于q>-1且q≠0可知Sn>0
bn=a[n+1]-ka[n+2]=a1q^n(1-kq)则{bn}也是等比数列,公比为q
且b1=a2-ka3=a1q(1-kq)
则Tn=a1q(1-kq)(1-q^n)/(1-q)
又Tn>kSn对于一切n∈N及满足条件的所有q都成立,
即a1q(1-kq)(1-q^n)/(1-q)>ka1(1-q^n)/(1-q),
得k
因为an为等比数列，所以可知：an=a1q^(n-1) Sn=a1(1-q^n)/(1-q)
又因为bn=a1q^n-ka1q^(n+1)=(1-kq)a1q^n所以bn也为等比数列，公比q，b1=q(1-kq)a1 Tn=b1(1-q^n)/(1-q)=q(1-kq)a1(1-q^n)/(1-q)
Tn>kSn
q(1-kq)a1(1-q^n)/(1-q)-k...
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因为an为等比数列，所以可知：an=a1q^(n-1) Sn=a1(1-q^n)/(1-q)
又因为bn=a1q^n-ka1q^(n+1)=(1-kq)a1q^n所以bn也为等比数列，公比q，b1=q(1-kq)a1 Tn=b1(1-q^n)/(1-q)=q(1-kq)a1(1-q^n)/(1-q)
Tn>kSn
q(1-kq)a1(1-q^n)/(1-q)-ka1(1-q^n)/(1-q)>0
(1-q^n)[q-k(q+1)]/(1-q)>0
即：(1-q)(1-q^n)[q-k(q+1)]>0
(1)q≠1时，
上式等价于：q-k(q+1)]>0
k(2)q=1时原式不成立
所以k由于q/(1+q)=1-1/(1+q)<1且≠0
故k<0或0
收起
an=a1q^(n-1)
Sn=a1(1-q^n)/(1-q)
bn=a1q^n-ka1q^(n+1)=(1-kq)a1q^n
故bn也为等比数列，公比q，b1=q(1-kq)a1
Tn=b1(1-q^n)/(1-q)=q(1-kq)a1(1-q^n)/(1-q)
Tn>kSn
q(1-kq)a1(1-q^n)/(1-q)-ka1(1-q^n)/(1...
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an=a1q^(n-1)
Sn=a1(1-q^n)/(1-q)
bn=a1q^n-ka1q^(n+1)=(1-kq)a1q^n
故bn也为等比数列，公比q，b1=q(1-kq)a1
Tn=b1(1-q^n)/(1-q)=q(1-kq)a1(1-q^n)/(1-q)
Tn>kSn
q(1-kq)a1(1-q^n)/(1-q)-ka1(1-q^n)/(1-q)>0
(1-q^n)[q-k(q+1)]/(1-q)>0
即：(1-q)(1-q^n)[q-k(q+1)]>0
(1)q≠1时，
上式等价于：q-k(q+1)]>0
k(2)q=1时原式不成立
所以k由于q/(1+q)=1-1/(1+q)<1且≠0
故k<0或0
收起