Question

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

Answer 1

当公比是1时,是常数列,an=a1

sn=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)*a1

a1=0

不成立

当公比不是一时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

2Sn=S(n+1)+S(n+2)

2a1(1-q^n)/1-q==a1(1-q^n+1)/1-q +a1(1-q^n+2)/1-q

2a1(1-q^n)=a1(1-q^n+1)+a1(1-q^n+2)

2-2q^n)=2-q^n+1-q^n+2

2q^n=q^n+1+q^n+2

解方程即可

2=q+q2

q=-2或q=1（舍）