Question

设数列{an}的前n项和为Sn=(m+1)－m*an,对任意正整数n都成立,m为小于－1的常数.

Answer 1

证明：Sn=(m+1)-m*an S(n+1)=(m+1)-m*a(n+1) ∴S(n+1)-Sn=a(n+1) =m*an-m*a(n+1) ∴a(n+1)/an= m/(m+1) ∵m＜-1,∴m/(m+1)≠0的常数 ∴｛an｝是等比数列.