设sn为数列{an}的前n项和对任意的n属于N*,都有Sn=(m+1)-man(m为常数,且m大于0）.

(1).证：Sn=(m+1)-man Sn-1=(m+1)-ma(n-1) an=Sn-Sn-1=(m+1)-man-(m+1)+ma(n-1) (m+1)an=ma(n-1) an/a(n-1)=m/(m+1) m为常数,且m>0,分数有意义,an/a(n-1)为常数.令n=1 a1=S1=(m+1)-ma1 (1+m)a1=m+1a1=1 数列{an}为等比数列,首项为1,公比为m/(m+1).(2).q=f(m)=m/(m+1) b1=2a1=2 bn=b(n-1)/[b(n-1)+1] b2=b1/(b1+1)=2/3 b3=b2/(b2+1)=(2/3)/(2/3+1)=2/5 假设n=k时,bk=2/(2k-1),则当n=k+1时 b(k+1)=bk/(bk+1) =[2/(2k-1)]/[2/(2k-1)+1] =2/[2+(2k-1)] =2/(2k+1) =2/[2(k+1)-1] 仍然满足同样的表达式 bn=2/(2n-1)