等比数列{an}的前n项和为Sn,已知S1,S3,S2成等差数列,求{an}的公比qS1+S2=2S3即a1+a1+a2=2(a1+a2+a3)解得2q^2+q=0,q=-1/2S1+S2=2S3即2a1(1-q^3)/(1-q)=a1(1-q)/(1-q)+a1(1-q^2)/(1-q)解得2q^2-q-1=0,q1=1,q2=-1/2为什么两个结果不
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![等比数列{an}的前n项和为Sn,已知S1,S3,S2成等差数列,求{an}的公比qS1+S2=2S3即a1+a1+a2=2(a1+a2+a3)解得2q^2+q=0,q=-1/2S1+S2=2S3即2a1(1-q^3)/(1-q)=a1(1-q)/(1-q)+a1(1-q^2)/(1-q)解得2q^2-q-1=0,q1=1,q2=-1/2为什么两个结果不](/uploads/image/z/5123787-51-7.jpg?t=%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E5%B7%B2%E7%9F%A5S1%2CS3%2CS2%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E6%B1%82%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%85%AC%E6%AF%94qS1%2BS2%3D2S3%E5%8D%B3a1%2Ba1%2Ba2%3D2%28a1%2Ba2%2Ba3%29%E8%A7%A3%E5%BE%972q%5E2%2Bq%3D0%2Cq%3D-1%2F2S1%2BS2%3D2S3%E5%8D%B32a1%281-q%5E3%29%2F%281-q%29%3Da1%281-q%29%2F%281-q%29%2Ba1%281-q%5E2%29%2F%281-q%29%E8%A7%A3%E5%BE%972q%5E2-q-1%3D0%2Cq1%3D1%2Cq2%3D-1%2F2%E4%B8%BA%E4%BB%80%E4%B9%88%E4%B8%A4%E4%B8%AA%E7%BB%93%E6%9E%9C%E4%B8%8D)
等比数列{an}的前n项和为Sn,已知S1,S3,S2成等差数列,求{an}的公比qS1+S2=2S3即a1+a1+a2=2(a1+a2+a3)解得2q^2+q=0,q=-1/2S1+S2=2S3即2a1(1-q^3)/(1-q)=a1(1-q)/(1-q)+a1(1-q^2)/(1-q)解得2q^2-q-1=0,q1=1,q2=-1/2为什么两个结果不
等比数列{an}的前n项和为Sn,已知S1,S3,S2成等差数列,求{an}的公比q
S1+S2=2S3
即a1+a1+a2=2(a1+a2+a3)
解得2q^2+q=0,q=-1/2
S1+S2=2S3
即2a1(1-q^3)/(1-q)=a1(1-q)/(1-q)+a1(1-q^2)/(1-q)
解得2q^2-q-1=0,q1=1,q2=-1/2
为什么两个结果不一样?
等比数列{an}的前n项和为Sn,已知S1,S3,S2成等差数列,求{an}的公比qS1+S2=2S3即a1+a1+a2=2(a1+a2+a3)解得2q^2+q=0,q=-1/2S1+S2=2S3即2a1(1-q^3)/(1-q)=a1(1-q)/(1-q)+a1(1-q^2)/(1-q)解得2q^2-q-1=0,q1=1,q2=-1/2为什么两个结果不
S1+S2=2S3
即a1+a1+a2=2(a1+a2+a3)
解得2q^2+q=0,q=-1/2
等比数列求和公式分2类,
一类公比q=1,Sn=na₁
第2类公比q≠1,Sn=a₁(1-qⁿ)/(1-q)
当q=1时,
S1+S2=2S3 ==>a1+2a1=2*3a1
==>a1=0与等比数列矛盾
当q≠1时,
即2a1(1-q^3)/(1-q)=a1(1-q)/(1-q)+a1(1-q^2)/(1-q)
解得2q^2-q-1=0,q1=1,q2=-1/2 (q1=1舍去)
∴q=-1/2
结果一样