已知数列{an}满足a1=2,a(n+1)=(3an-2)/(2an-1),求证{1/(an-1)}是等差数列,并求数列{an}的通项公式
人气:351 ℃ 时间:2019-11-24 11:04:02
解答
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...3-1/[2(2an-1)]这一步是什么意思
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