Question

已知数列{an}中.a1=3/5.an=2-1/[a(n-1)].数列{bn}满足bn=1/(an-1)..求数列bn为等差数列

已知数列{an}中.a1=3/5.an=2-1/[a(n-1)].数列{bn}满足bn=1/(an-1)..求数列bn为等差数列.（2）数列an的最大项和最小项

Answer 1

b(1)=1/(a(1)-1)=-5/2b(n)-b(n-1)=1/(a(n)-1)-1/(a(n-1)-1)=1/(2-1/a(n-1)-1)-1/(a(n-1)-1)=a(n-1)/(a(n-1)-1)-1/(a(n-1)-1)=1b(n)=n-7/2是等差数列a(n)=1+a/(n-7/2)a(1)=3/5a(2)=1/3a(3)=-1a(4)=31<a(n)<3，当n>4时所以a(4)=3最大项，a(3)=-1最小项