Question

已知数列{An}的通项公式是An=n/n^2+156(n属于正整数)则数列的最大项是多少

Answer 1

a(n)=n/[n^2 + 156] = 1/[n + 156/n] <= 1/[ 2(n*156/n)^(1/2)] = 1/伍吵腊[2(4*39)^(1/2)],
n = 156/腔滑n, n^2 = 156, 12<=n<=13.
a(12)=1/[12 + 156/12] = 1/[12 + 13] = 1/25,
a(13) =1/[13+156/13] = 1/[13+12] = 1/25,
数碰举列的最大项是a(12)=a(13)=1/25

追问

1/[n + 156/n] <= 1/[ 2(n*156/n)^(1/2)] = 1/[2(4*39)^(1/2)],
什么意思没看懂

追答

x为任意正数时，f(x)=1/[x+156/x] 0,b>0时，a+b>=2(a*b)^(1/2). 1/[a+b] <= 1/[2(ab)^(1/2)].

当x为正整数n时，a(n) = 1/[n+156/n] <= max[a(12),a(13)],
而，a(12)=a(13)=1/25,
因此，总有a(n) <= a(12) = a(13) = 1/25.