设等比数列{an}的公比为q,前n项和为Sn,若S(n+1),Sn,S(n+2)成等差数,则q的值为

问题描述:

设等比数列{an}的公比为q,前n项和为Sn,若S(n+1),Sn,S(n+2)成等差数,则q的值为

当公比是1时,是常数列,an=a1sn=n*a1,sn+1=(n+1)*a1,sn+2=(n+2)*a1若S(n+1),Sn,S(n+2)成等差2Sn=S(n+1)+S(n+2)2n*a1=(n+1)*a1+(n+2)*a1a1=0不成立当公比不是一时Sn=a1(1-q^n)/1-qSn+1=a1(1-q^n+1)/1-qSn+2=a1(1-q^n+2...