Question

数列{an}满足a(n+1)=3an-2/2an-1,且a1=2.（1）设bn=1/an-1,求证{bn}为等差数列.（

Answer 1

a(n+1)=(3an-2)/(2an-1)=(3an-3/2-1/2)/(2an-1)=3-1/[2(2an-1)]= →

a(n+1)=(3an-2)/(2an-1) → a(n+1)-1=(3an-2)/(2an-1)-1=(an-1)/(2an-1)

→ 1/[a(n+1)-1]=(2an-1)/(an-1)=1/(an-1)+2

∴1/(an-1)是公差为2的等差数列

1/(a1-1)=1/(2-1)=1,∴1/(an-1)=1+2(n-1)=2n-1

→ (an-1)=1/(2n-1)

→an=1+1/(2n-1)=2n/(2n-1)