设等比数列{an}的公比为q,前n项和为Sn,若Sn+1,Sn,Sn+2成等差数列,则q=?答案是-2或1
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![设等比数列{an}的公比为q,前n项和为Sn,若Sn+1,Sn,Sn+2成等差数列,则q=?答案是-2或1](/uploads/image/z/2718767-47-7.jpg?t=%E8%AE%BE%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%85%AC%E6%AF%94%E4%B8%BAq%2C%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E8%8B%A5Sn%2B1%2CSn%2CSn%2B2%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%88%99q%3D%3F%E7%AD%94%E6%A1%88%E6%98%AF-2%E6%88%961)
设等比数列{an}的公比为q,前n项和为Sn,若Sn+1,Sn,Sn+2成等差数列,则q=?答案是-2或1
设等比数列{an}的公比为q,前n项和为Sn,若Sn+1,Sn,Sn+2成等差数列,则q=?
答案是-2或1
设等比数列{an}的公比为q,前n项和为Sn,若Sn+1,Sn,Sn+2成等差数列,则q=?答案是-2或1
因为Sn+1,Sn,Sn+2成等差数列
S(n+1)+S(n+2)=2*S(n)
(q^(n+1)-1)*a1/(q-1)+(q^(n+2)-1)*a1/(q-1)=2*(q^(n)-1)*a1/(q-1)
q^(n+1)-1+q^(n+2)-1=2*q^n-2
q*q+q-2=0
所以q=-2或1
若q=1,则Sn=na1,式显然不成立
所以q=-2
解:1设等比数列{an}的公比为q=1,首项为a1有:通项an=a1,前n项和为Sn=na1,Sn+1=(n+1)a1 Sn+2=(n+2)a1 则有2Sn=Sn+2+Sn+2 有Sn+1,Sn,Sn+2成等差数列满足条件.
2. 设等比数列{an}的公比为q(q≠1,)首项为a1有:通项an=a1*q^(n-1), Sn=[a1*(1-q^n)]/(1-q) 其中q≠1 ...
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解:1设等比数列{an}的公比为q=1,首项为a1有:通项an=a1,前n项和为Sn=na1,Sn+1=(n+1)a1 Sn+2=(n+2)a1 则有2Sn=Sn+2+Sn+2 有Sn+1,Sn,Sn+2成等差数列满足条件.
2. 设等比数列{an}的公比为q(q≠1,)首项为a1有:通项an=a1*q^(n-1), Sn=[a1*(1-q^n)]/(1-q) 其中q≠1 又因为Sn+1,Sn,Sn+2成等差数列,则Sn-Sn+1=S(n+2)-Sn有:a1*(1-q^n)-a1*(1-q^n+1)=a1*(1-q^n+2)-a1*(1-q^n)化简可得:1+q=-1 故有q=-2
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