多情的眼睛
幼苗
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运用数学归纳法!
比较 x = k^(k+1) 和 y = (k+1)^k的大小
x(1) = 1 < y(1) = 2
x(2) = 8 < y(1) = 9
x(3) = 81 > y(1) = 64
x(4) = 1024 > y(1) = 625
那么假设 x(k) > y(k), k>=3 成立,即 k^(k+1) > (k+1)^k , 那么
(k+1)^(k+2) - (k+2)^(k+1)
= [ k * (k+1)/k ]^(k+1) * (k+1) - (k+2)^(k+1)
= k^(k+1) * [(k+1)/k ]^(k+1) * (k+1) - (k+2)^(k+1)
> (k+1)^k * [(k+1)/k ]^(k+1) * (k+1) - (k+2)^(k+1)
= (k+1)^(2k+2) / k^(k+1) - (k+2)^(k+1)
= [ (k+1)^2 ]^(k+1) / k^(k+1) - (k+2)^(k+1)
= [ (k+1)^2 / k ]^(k+1) - (k+2)^(k+1)
= (k + 2 + 1/k)^(k+1) - (k+2)^(k+1)
> 0
所以 当 k>=3 的时候, k^(k+1) > (k+1)^k 成立!
k=8的时候, 8^9 > 9^8
1年前
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