a(n+1)=2an-a(n-1) 3bn-b(n-1)=n

题目

a(n+1)=2an-a(n-1) 3bn-b(n-1)=n
数列{an},a(n+1)=2an-a(n-1),a1=1/4,a2=3/4.数列{bn},3bn-b(n-1)=n,{bn}前n项和Sn
1.求证数列{bn-an}是等比
2.若当且仅当n=4时,Sn取得最小值,求b1取值范围

答案

1.a(n+1)=2an-a(n-1)a(n+1)-an=an-a(n-1)an为以1/4为首项,1/2为公差的等差数列an=n/2-1/4bn-an=bn-n/2+1/4b(n+1)-a(n+1)=bn/3+n/3+1/3-n/2-1/2+1/4=bn/3-n/6+1/12=(bn-an)/3得证2.b4=(b1+49)/270-184