已知{An}是以a为首项,q为公比的等比数列,Sn为它的前n项和 求当Sm;Sn;Sk成等差数列时,求证:对任意自然数k
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已知{An}是以a为首项,q为公比的等比数列,Sn为它的前n项和 求当Sm;Sn;Sk成等差数列时,求证:对任意自然数k,
已知{An}是以a为首项,q为公比的等比数列,Sn为它的前n项和 当Sm;Sn;Sl成等差数列时,求证:对任意自然数k,Am+k,An+k,Al+k也成等差数列.
已知{An}是以a为首项,q为公比的等比数列,Sn为它的前n项和 当Sm;Sn;Sl成等差数列时,求证:对任意自然数k,Am+k,An+k,Al+k也成等差数列.
![已知{An}是以a为首项,q为公比的等比数列,Sn为它的前n项和 求当Sm;Sn;Sk成等差数列时,求证:对任意自然数k](/uploads/image/z/4609873-1-3.jpg?t=%E5%B7%B2%E7%9F%A5%7BAn%7D%E6%98%AF%E4%BB%A5a%E4%B8%BA%E9%A6%96%E9%A1%B9%2Cq%E4%B8%BA%E5%85%AC%E6%AF%94%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2CSn%E4%B8%BA%E5%AE%83%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C+%E6%B1%82%E5%BD%93Sm%3BSn%3BSk%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E6%97%B6%2C%E6%B1%82%E8%AF%81%3A%E5%AF%B9%E4%BB%BB%E6%84%8F%E8%87%AA%E7%84%B6%E6%95%B0k)
An=aq^(n-1)
Sn=a(q^n -1)/(q-1)
Sm=a(q^m -1)/(q-1)
Sl=a(q^l -1)/(q-1)
∵2Sn=Sm+Sl,代入化简得:2q^n=q^m+q^l
A(n+k)=aq^(n+k-1)=(q^n)*aq^(k-1)
A(m+k)=(q^m)*aq^(k-1)
A(l+k)=(q^l)*aq^(k-1)
∴2A(n+k)=A(m+k)+A(l+k)
Sn=a(q^n -1)/(q-1)
Sm=a(q^m -1)/(q-1)
Sl=a(q^l -1)/(q-1)
∵2Sn=Sm+Sl,代入化简得:2q^n=q^m+q^l
A(n+k)=aq^(n+k-1)=(q^n)*aq^(k-1)
A(m+k)=(q^m)*aq^(k-1)
A(l+k)=(q^l)*aq^(k-1)
∴2A(n+k)=A(m+k)+A(l+k)
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