设公比为q.
q=1时,Sm=ma1 S(m+2)=(m+2)a1 S(m+1)=(m+1)a1
2S(m+2)=2(m+2)a1=(2m+4)a1
Sm+S(m+1)=ma1+(m+1)a1=(2m+1)a1
数列为等比数列,a1≠0,又2m+4≠2m+1,2S(m+2)≠Sm+S(m+1),与题意不符,因此q≠1.
这步判断一定要有.
2S(m+2)=Sm +S(m+1)
2a1[q^(m+2) -1]/(q-1)=a1(q^m -1)/(q-1) +a1[q^(m+1)-1]/(q-1)
整理,得
2q^(m+2)=q^m +q^(m+1)
2q^2-q-1=0
(q-1)(2q+1)=0
q=1(舍去)或q=-1/2
a(m+2)=a1q^(m+1)=(-1/2)a1q^m
am=a1q^(m-1)=a1q^m/q=(-2)a1q^m
a(m+1)=a1q^m
2a(m+2)=-a1q^m
am+a(m+1)=(-2)a1q^m+a1q^m=-a1q^m
2a(m+2)=am+a(m+1)
am,a(m+2),a(m+1)成等差数列.