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

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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