Question

在数列中a1=1,3anan-1+an-an-1=0(n≥2,n∈N)若入an+1/an+1≥入对任意n≥2的整数恒成立

Answer 1

3anan-1+an-an-1=0两边同除ana(n-1)得：3+1/a(n-1)-1/an=0 1/an-1/a(n-1)=3 1/a1=1

所以,{1/an}是以1为首项、以3为公差的等差数列,1/an=3n-2 an=1/(3n-2)(n=1,2,…)

an+1/a(n+1)=1/(3n-2)+3n+1=(9n^2-3n-1)/(3n-2)=[(3n-2)^2+3(3n-2)+1]/(3n-2)=(3n-2)+1/(3n-2)+3

(3n-2)+1/(3n-2)>=2,当(3n-2)=1/(3n-2)即n=1时取等号.当n>=2时,an+1/a(n+1)递增.

所以,当n=2时,an+1/a(n+1)取最小值29/4

所以,入