Question

数列an中,a1=五分之三,an=2-an-1分之一,数列bn满足bn=bn-1分之一

已知数列{an}中,a1=3/5,通项an=2-1/a(n-1)（n≥2,n∈N*）,数列{bn}满足bn=1/(an-1)(n∈N*)
证明数列{bn}是等差数列,并求出通项bn

Answer 1

an=2-1/a(n-1),ana(n-1)=2a(n-1)-1,an-1=1-1/a(n-1),1/(an-1)=a(n-1)/[a(n-1)-1],则bn=1/(an-1)=a(n-1)/[a(n-1)-1],b(n+1)=an/(an-1),b(n+1)-bn=an/(an-1)-a(n-1)/[a(n-1)-1]
=[ana(n-1)-an-ana(n-1)+a(n-1)]/[ana(n-1)-an-a(n-1)+1]=[a(n-1)-an]/[2a(n-1)-1-an-a(n-1)+1]=1,则b(n+1)-bn=1,数列{bn}是等差数列,公差d=1,b1=1/(3/5-1)=-5/2,bn=-5/2+n-1=n-7/2
通项bn=n-7/2