等比数列AN公比为Q前N项和为SN,若Sn+1,Sn,Sn+2成等差数列,求Q
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![等比数列AN公比为Q前N项和为SN,若Sn+1,Sn,Sn+2成等差数列,求Q](/uploads/image/z/1770530-50-0.jpg?t=%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97AN%E5%85%AC%E6%AF%94%E4%B8%BAQ%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%E6%B1%82Q)
等比数列AN公比为Q前N项和为SN,若Sn+1,Sn,Sn+2成等差数列,求Q
等比数列AN公比为Q前N项和为SN,若Sn+1,Sn,Sn+2成等差数列,求Q
等比数列AN公比为Q前N项和为SN,若Sn+1,Sn,Sn+2成等差数列,求Q
公比为1时显然不成立.故有:
Sn=a1(1-q^n)/(1-q)
Sn+1=a1[1-q^(n+1)]/(1-q)
Sn+2=a1[1-q^(n+2)]/(1-q)
Sn+1,Sn,Sn+2成等差数列,故有
Sn+1 + Sn+2=2Sn
即2a1(1-q^n)/(1-q)=a1[1-q^(n+1)]/(1-q)+a1[1-q^(n+2)]/(1-q)
化简得 2-2q^n=2-q^(n+1)-q^(n+2)
q^2+q-2=0
解出q=-2
(1+根号5)/2 或(1-根号5)/2