培正數學邀請賽

20 (Suggested Solution)

As lim(x -> 0) {1 - [cos(kx)]^(1/k)}/x^2 = k/2

The equivalent form is that when x -> 0

cos(kx)^(1/k) = 1 - kx^2/2

So, cos(x) * [cos(2x)]^(1/2) * [cos(3x)]^(1/3) * ... * [cos(100x)]^(1/100)

= (1 - x^2/2)(1 - 2x^2/2)(1 - 2x^2/2)...(1 - 100x^2/2)

= 1 - (x^2/2 + 2x^2/2 + ... + 100x^2/2) + higher terms of x

lim(x -> 0) {1 - cos(x) * [cos(2x)]^(1/2) * [cos(3x)]^(1/3) * ... * [cos(100x)]^(1/100)}/x^2

= lim(x -> 0) {1 - [1 - (x^2/2 + 2x^2/2 + ... + 100x^2/2)]}/x^2

= (1 + 2 + ... + 100)/2

= 2025





2013-01-18 21:38:23 補充：
不知拙作是不是閣下心目中的解題方法

2013-01-18 21:41:58 補充：
(1 + 2 + ... + 100)/2 = 2525

myisland8132 - I am not sure if your answer is rigorous.

2013-01-23 02:23:41 補充：
http://s187.beta.photobucket.com/user/cshung/media/7013011800007_zpsdcd2161a.png.html

f(n)=n/2+ f(n-1), f(1)= 1/2
so f(n)= n/2 + (n-1)/2 +...+2/2 + f(1)= (1+2+...+n)/2 = n(n+1)/2

想不到怎用HINT , 用最基本的方法做 :

圖片參考：http://imgcld.yimg.com/8/n/HA00482599/o/20130118120049.jpg

答案是 2525，暫時沒空，今晚有機會回答你的問題。