已知函数f(x)=(x^2)/2x-2,如果数列{an}满足a1=4,a(n+1)=f(an),求证:当n>=2时,恒有an<3成立.
人气:314 ℃ 时间:2020-04-28 18:21:30
解答
a(n+1)=f(an)=an^2 /(2an -2)
a(n+1) -2=(an^2 -4an +4)/[2(an -1)]=(an -2)^2 /[2(an -1)]
a1=4>1 a2-2=(a1-2)^2/[2(a1 -1)]=2/3>0,a2=8/3
2
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