數學題X1

在數學版問怎麼用 英文 說 台灣迷信 ？

2009-08-08 12:41:41 補充：
我還在想怎麼用一個式子表示An

2009-08-08 13:49:14 補充：
答案: An=[1-{(n+2)/4 - [(n+2)/4]}]*2 + [1-{n/4 - [n/4]}]*4
(高斯符號 [x]表示不大於x的最大整數，如[3.4]=3，[5]=5)

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[說明]
1.當n=4k+1(其中k為非負整數)時：
(n+2)/4 - [(n+2)/4] = (k + 3/4) - [k + 3/4] = (k + 3/4) - k = 3/4
n/4 - [n/4] = (k + 1/4) - [k + 1/4] = (k + 1/4) - k = 1/4

=> [1-{(n+2)/4 - [(n+2)/4] }] = [1-3/4] = [1/4] = 0
　 [1-{n/4 - [n/4]}] = [1-1/4] = [3/4] = 0

=> An = 0*2 + 0*4 = 0

2.當n=4k+2(其中k為非負整數)時：
(n+2)/4 - [(n+2)/4] = (k + 4/4) - [k + 4/4] = (k + 1) - (k + 1) = 0
n/4 - [n/4] = (k + 2/4) - [k + 2/4] = (k + 1/4) - k = 1/4

=> [1-{(n+2)/4 - [(n+2)/4] }] = [1-0] = [1] = 1
　 [1-{n/4 - [n/4]}] = [1-1/4] = [3/4] = 0

=> An = 1*2 + 0*4 = 2

3. 當n=4k+3(其中k為非負整數)時：
(n+2)/4 - [(n+2)/4] = (k + 5/4) - [k + 5/4] = (k + 1 + 1/4) - (k+1) = 1/4
n/4 - [n/4] = (k + 3/4) - [k + 3/4] = (k + 3/4) - k = 3/4

=> [1-{(n+2)/4 - [(n+2)/4] }] = [1-1/4] = [3/4] = 0
　 [1-{n/4 - [n/4]}] = [1-3/4] = [1/4] = 0

=> An = 0*2 + 0*4 = 0

4. 當n=4k+4(其中k為非負整數)時：
(n+2)/4 - [(n+2)/4] = (k + 6/4) - [k + 6/4] = (k + 1 + 2/4) - (k + 1) = 2/4
n/4 - [n/4] = (k + 4/4) - [k + 4/4] = (k + 1) - (k + 1) = 0

=> [1-{(n+2)/4 - [(n+2)/4] }] = [1-2/4] = [2/4] = 0
　 [1-{n/4 - [n/4]}] = [1-0] = [1] = 1

=> An = 0*2 + 1*4 = 4


2009-08-09 00:20:54 補充：
真的有錯，哈，不好意思，改版太多次，有的地方沒改到。
目前在想能不能用sin來寫通式

(參考一下!)
0,2,0,4四個一循環,可設A(n)=a+bi^n+c(-1)^n+d(-i)^n , ( i^2 = -1)
A(1)=0 = a+bi- c - di
A(2)=2 = a- b + c - d
A(3)=0 = a - bi - c +di
A(4)=4 = a+ b + c + d
解四式聯立得 a=c= 3/2, b=d= 1/2
故A(n)=[ 3+i^n+ 3(-1)^n +(-i)^n ] / 2, n=1,2,3,4,....

答案可以有很多型式

n=2k+1或n=2k+3時，An=0
n=2k，k屬於奇數時，An=2
k屬於偶數時，An=4

n=4k時：An=4

n=4k+1：An=0

n=4k+2：An=2

n=4k+3：An=0

幫忙一下
很急
卸卸

http://tw.knowledge.yahoo.com/question/question?qid=1509080703394