Question

已知数列{an}满足a1=100，an+1-an=2n 则求an/n的最小值?

已知数列{an}满足a1=100，an+1-an=2n 则求an/n的最小值?

Answer 1

a(2) - a(1) = 2
a(3) - a(2) = 4
......
a(n+1) - a(n) = 2n
这n个式子相加，就有
a(n+1) = 100+ n(n+1)
即a(n) = n(n-1) +100 = n^2 - n + 100
a(n)/n = n + 100/n -1
用均值不等式，知道它在n = 10的时候取最小值19

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Answer 2

a(n+1)-a1=(2+2n)*n/2=n*(n+1)
a(n+1)=n*(n+1)+100
a(n)=(n-1)*n+100
a(n)/n=n-1+100/n
当n=10时有最小值19