有奖解题

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1 | 2007-9-19 23:43:28 | 显示全部楼层 |阅读模式

f(n+1) - f(n)=f(1) +2(n+1) + 1 = 2n + 4 >0所以f(x)在x∈Z+上严格递增所以当x≥2时,f(x)≥f(2) = f(1+1) = 1 + 1 + 2(1+1) +1 = 7 ≥ (a+7)*x-(a+10)即(a+7)x-(a+7) - 3 ≤7(a+7)(x-1) ≤ 10∵x ≥ 2 ∴x-1 ≥ 1 > 0 ∴1/(x-1) ≤ 1∴(a+7) ≤ 10/(x-1)∴a ≤ 10/(x-1) - 7 ≤ 10/1 - 7 = 3

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