✔ 最佳答案
For all n>0,(n+1)/(n+2)-n/(n+1)=[(n+1)^2-n(n+2)]/[(n+1)(n+2)]=1/[(n+1)(n+2)]>0∴ (n+1)/(n+2)>n/(n+1) for all n>0 ⋯⋯ (*)
P(n):n^(n+1)/(n+1)^n>1 for all integers n>2When n=3,LHS=3^4 / 4^3=81/64>1=RHS∴ P(3) is true.Assume P(k) is true, ie.k^(k+1)/(k+1)^k>1When n=k+1LHS=(k+1)^(k+2)/(k+2)^(k+1)=(k+1)*(k+1)^(k+1)/(k+2)^(k+1)=(k+1)*[(k+1)/(k+2)]^(k+1)>(k+1)*[k/(k+1)]^(k+1) ⋯⋯ (from (*))=k^(k+1)/(k+1)^k>1So, n=k+1 is also true.According to MI, P(n) is true for all integer greater than 2.
It can be deducted thatn^(n+1)/(n+1)^n>1==> n^(n+1)>(n+1)^n==> n^(n+1)-(n+1)^n>0 for n greater than 2So, it is not true for n=1 or 2.When n=1,1^2-2^1=1-2=-1,When n=2,2^3-3^2=8-9=-1.So, if | a^(a+1)-(a+1)^a |=1 and a is a positive integer,then a=1 or 2.
2014-10-22 09:38:19 補充:
P(n) 已經指出 當 n 大於 2 時,n^(n+1)/(n+1)^n 的比例大於1,
即 當 n 大於 2 時,n^(n+1)-(n+1)^n 永遠大於1。
所以對於你這條題目,是無需考慮 n>2 的。
2014-10-22 12:32:45 補充:
對不起,是
當 n 大於 2 時,n^(n+1)-(n+1)^n 永遠大於 0。
而當 n 大於 2 時,n^(n+1) 與 (n+1)^n 的相差是越多越大的。
2014-10-22 20:17:17 補充:
P(n) 不是利用咗 (*) 來證明了嗎?
2014-10-22 23:56:36 補充:
LHS
=(k+1)^(k+2)/(k+2)^(k+1)
=(k+1)*(k+1)^(k+1)/(k+2)^(k+1)
=(k+1)*[(k+1)/(k+2)]^(k+1)
>(k+1)*[k/(k+1)]^(k+1) ⋯⋯ (from (*))
=k^(k+1)/(k+1)^k
>1
所以 n^(n+1) 與 (n+1)^n 的相差是越多越大的。