已知数列{an}满足a1=100,an+1-an=2n,则a
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已知数列{an}满足a1=100,an+1-an=2n,则
a
a2-a1=2,
a3-a2=4, … an+1-an=2n, 这n个式子相加,就有an+1=100+n(n+1), 即an=n(n-1)+100=n2-n+100, ∴ an n=n+ 100 n-1≥2 n• 100 n-1=19, 当且仅当n= 100 n,即n=10时, an n取最小值19. 故答案为:19.
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