n^(n+1)与(n+1)^n大小 归纳法

2025-02-27 00:48:53
推荐回答(1个)
回答1:

你给的答案我也看不懂,我另给答案吧。
解:当n=1时,1^2<2^1,n^(n+1)<(n+1)^n
当n=2时,2^3<3^2,n^(n+1)<(n+1)^n
当n=3时,3^4>4^3,n^(n+1)>(n+1)^n
设当n=k(k>=3)时,k^(k+1)>(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。