prove that
(x^(k+1)-1)/(k+1) [>or=] (x^k-1)/k
x belongs to positive real numbers
k belongs to positive integers
pure maths inequality
2008-03-23 6:37 pm
回答 (2)
2008-03-23 7:32 pm
✔ 最佳答案
The results are as follows~~~圖片參考：http://i182.photobucket.com/albums/x4/A_Hepburn_1990/A_Hepburn02Mar231131.jpg?t=1206243129
參考: Myself~~~
2008-03-23 7:28 pm
Since
x^p-1＞p(x-1) for [x,p positive]
Now substitute x=x^k, p=(k+1)/k
x^p-1＞p(x-1)
x^(k+1)-1＞[(k+1)/k](x^k-1)
[x^(k+1)-1]/(k+1)＞(x^k-1)/k
2008-03-23 11:44:08 補充：
我在上面大於號後面打漏了個等於號.......
x^p-1＞p(x-1) for [x,p positive]
Now substitute x=x^k, p=(k+1)/k
x^p-1＞p(x-1)
x^(k+1)-1＞[(k+1)/k](x^k-1)
[x^(k+1)-1]/(k+1)＞(x^k-1)/k
2008-03-23 11:44:08 補充：
我在上面大於號後面打漏了個等於號.......
參考: Introduction to Elliptic Curves and Modular Forms (Graduate Texts in Mathematics) (Hardcover) by Neal I. Koblitz
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