1.已知an=n/n^2+156(n∈N),则在数列{an}的最大项为2.数列{an}的通项为an=an/bn+a,其中a,b均为正数,则an与an+1的大小关系为

Answer 1

an=n/(n^2+156)
an-an-1=[n(n-1)^2+156n-(n-1)n^2-156(n-1)]/[(n^2+156)((n-1)^2-156)]
=(156+n^3-2n^2+n-n^3+n^2)/[(n^2+156)((n-1)^2-156)]
=(156-n^2+n)/[(n^2+156)((n-1)^2-156)]
156-n^2+n>=0
(n-1/2)^2