已知数列{an}中,a1=12,an+1－an=2n,则an/n的最小值为?

Question

已知数列{an}中,a1=12,an+1－an=2n,则an/n的最小值为?

Answer 1

a(n+1) = a(n) + 2n = a(n) + n(n+1) - (n-1)n,
a(n+1)-n(n+1) = a(n)-(n-1)n,
{a(n)-(n-1)n}是首项为a(1)=12,的常数数列.
a(n) - (n-1)n = 12,
a(n) = 12 + (n-1)n.
a(n)/n = 12/n + n -1,
a(n+1)/(n+1) = 12/(n+1) + n,
a(n+1)/(n+1) - a(n)/n = 12/(n+1) - 12/n + 1 = [n(n+1) - 12]/[n(n+1)]
= (n+4)(n-3)/[n(n+1)],
1= a(4)/4 = 6.
因此,总有,a(n)/n >= a(4)/4 = 6.
a(n)/n的最小值为a(4)/4 = 6.