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的值为
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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的值为
![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的值为](/uploads/image/z/15928116-60-6.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%E7%9A%84%E5%80%BC%E4%B8%BA)
2Sn=S(n+2)+S(n+1)(q不为1)
2a1(1-q^n)/(1-q)=a1[1-q^(n+2)]/(1-q)+a1[1-q^(n+1)]/(1-q)
2-2q^n=2-[q^(n+1)+q^(n+2)]
那么q+q^2=0
可得q=-1或q=0(舍去)
当q=1,Sn=na1
Sn+1=(n+1)a1
S(n+2)=(n+2)a1
显然a1=0,.
那就不可以了.
所以q=-1
2a1(1-q^n)/(1-q)=a1[1-q^(n+2)]/(1-q)+a1[1-q^(n+1)]/(1-q)
2-2q^n=2-[q^(n+1)+q^(n+2)]
那么q+q^2=0
可得q=-1或q=0(舍去)
当q=1,Sn=na1
Sn+1=(n+1)a1
S(n+2)=(n+2)a1
显然a1=0,.
那就不可以了.
所以q=-1
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