高一数学
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 (1)q=1时,{an}为常数列,Sn=na1, 显然满足条件 (2)q≠1时:Sn+1,Sn,Sn+2成等差数列--->2Sn = S(n-1)+S(n+1) --->2a1(q^n-1) = a1[q^(n+1)-1] + a1[q^(n+2)-1] --->2q^n = q^(n+1) + q^(n+2) --->q²+q-2 = 0 --->(q-1)(q+2)=0--->q=-2, 综合(1)(2),q=1或q=-2
Sn、Sn+1、Sn+2成等差数列,故Sn+2 - Sn+1=Sn+1 - Sn;而Sn+2 - Sn+1=an+2=a1q^(n+1),Sn+1 - Sn=an+1=a1q^n,故有a1q^(n+1)=a1q^n ==> a1q^n(q-1)=0,a1q^n不=0,故q=1。可见,{an}又是一个常数列。
答:第1小题,解: S5=a1*(1-q^5)/(1-q) ; S15=a1*(1-q^15)/(1-q) S10=a1*(1-q^10)/(1-q) ;S20=a...详情>>
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