n^(n+1)与(n+1)^n大小 归纳法
1个回答

你给的答案我也看不懂,我另给答案吧.

当n=1时,1^2(k+1)^k,即

k^(k+1)/(k+1)^k>1

k*(k/(k+1))^k>1

当n=k+1时,考察(k+1)^(k+2)>(k+2)^(k+1)是否成立.

∵k^2+2k+1>k^2+2k

∴(k+1)^2>k(k+2)

(k+1)^2/(k+2)>k

(k+1)/(k+2)>k/(k+1)

((k+1)/(k+2))^k>(k/(k+1))^k

k*((k+1)/(k+2))^k>k*(k/(k+1))^k>1

(k+1)^2/(k+2)*((k+1)/(k+2))^k>k*((k+1)/(k+2))^k>1

(k+1)^(k+2)/(k+2)^(k+1)>1

(k+1)^(k+2)>(k+2)^(k+1)

根据数学归纳法,当n>=3时,n^(n+1)>(n+1)^n成立.

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