数列{an}中,a1=1/5,且当n>=2时,(an-1)/an=[(2an-1)+1]/1-2an,求数列的通项公已知数列{an}满足a1=1/5,且当n>1,n∈N*时,有a(n-1)/an=2a(n-1)+1/1-2an,求数列的通项公式an

Question

数列{an}中,a1=1/5,且当n>=2时,(a n-1)/an=[(2a n-1)+1]/1-2an,求数列的通项公
已知数列{an}满足a1=1/5,且当n>1,n∈N*时,有a(n-1)/an=2a(n-1)+1/1-2an,求数列的通项公式an

Answer 1

∵a(n-1)/an=[2a(n-1)+1[/[1-2an]
∴an-1－2anan-1＝2anan-1＋an ∴an-1－an＝4anan-1 ∴1/an－1/an-1＝4
∴数列{1/an} 是首项为1/a1＝5,公差为4的等差数列
∴1/an ＝5＋4(n－1) ＝4n＋1
∴an＝1/(4n＋1)