| 發布者 | 內容列 |
r1742006
Just popping in
註冊日: 2006-10-23
發表數: 1
|  證明題 |  | 1.1*3*5*7*9.......*2005)/1004*1005*1006........*2006=？
2.[1/(1*2)+1/(3*4)+1/(5*6).......+1/(2005*2006)]/
1/(1004*2006)+1/(1005*2005)+.......+1/(2006*1004)]
=
3.證：n的5次方減n 恆為30的倍數
4.證:Sm=Sn 則Sm+n=0 |
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 2006-11-29 21:27 |  |
訪客
| | Re: 證明題 | | 3.2帶入及矛盾 |
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| 2006-11-30 17:46 | |
s87009812
Just can't stay away
註冊日: 2004-07-25
發表數: 132
台北市立景美國中
| | Re: 證明題 | | 2.
[1/(1*2)+1/(3*4)+1/(5*6).......+1/(2005*2006)]=1-1/2+1/3-1/4+1/5-1/6+...+1/2005-1/2006=1+1/2+1/3+1/4+...+1/2006-2(1/2+1/4+...+1/2006)=1+1/2+1/3+...+1/2006-(1+1/2+1/3+...+1/1003)=1/1004+1/1005+...+1/2006
1/(1004*2006)+1/(1005*2005)+.......+1/(2006*1004)]
=2[1/(1004*2006)+1/(1005*2005)+...+1/(1504*1506)]+1/(1505*1505)=2*(1/3010)*[1/1004+1/2006+1/1005+1/2005+...1/1504+1/1506+1/1505]
兩式相除=1/(2/3010)=1505
_________________
想要得到什麼,就必須付出相同的代價...這就是煉金術中所說的"等價交換原則".
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| 2006-11-30 20:41 |  |
訪客
| | Re: 證明題 | | 你是內湖高中的嗎?
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| 2006-12-03 18:52 | |
訪客
| | Re: 證明題 | | 3.2^5-2=32-2=30
沒有矛盾啊! |
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| 2006-12-03 19:00 | |
s87009812
Just can't stay away
註冊日: 2004-07-25
發表數: 132
台北市立景美國中
| | Re: 證明題 | | 引文:
寫道:
3.2^5-2=32-2=30
沒有矛盾啊!
本題最大問題在於n必須為一自然數.(但你沒講)
n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=(n-1)n(n+1)(n^2+1).
(*連續三自然數必有一數為3的倍數,且至少有一數為偶數.)
當n=5k時 (n-1)n(n+1)(n^2+1)被30整除.
當n=5k+1時 n-1為5的倍數,又根據*的推論,命題成立.
當n=5k+2時 n^2+1=25k^2+20k+5為5的倍數,命題成立.
當n=5k+3時 n^2+1=25k^2+30k+10為5的倍數,命題成立.
當n=5k+4時 n+1=5k+5為5的倍數,命題成立.
綜上,n^5-n必被30整除. (n屬於N)
_________________
想要得到什麼,就必須付出相同的代價...這就是煉金術中所說的"等價交換原則".
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| 2006-12-03 21:10 | |
black
Just popping in
註冊日: 2007-01-14
發表數: 6
| | Re: 證明題 | | 1x3x5…x2003x2005/1004x1005x1006…x2006
=1x2x3x4…x2004x2005x2006/2x4…x2006x1004x1005x…x2006(擴分)
=1x2x3x4…x2004x2005x2006/（1x2)（2x2）（2x3）…（2x1003）x1004x1005…x2006
=1x2x3x4…x2004x2005x2006/(2的1003次方)1x2x3x4…x2004x2005x2006
=1/(2的1003次方)
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| 2007-02-05 15:55 | |
mathematical2005
Just can't stay away
註冊日: 2005-05-17
發表數: 117
China
| | Re: 證明題 | | 3.
n^5-n=(n-1)n(n+1)(n^2+1)
其中(n-1)n(n+1)為連續三正整數乘積
故必為2或3之倍數
n之尾數若為0,1,4,5,6,9則(n-1)n(n+1)必有一項為5之倍數
n之尾數若為2,3,7,8則(n^2+1)必為5之倍數
故n的5次方減n 恆為30(2*3*5)的倍數
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The art of doing mathematics consists in finding that special case which contains all the germs of generality 
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| 2007-08-03 18:00 | |
mathematical2005
Just can't stay away
註冊日: 2005-05-17
發表數: 117
China
| | Re: 證明題 | | 我有點兒看不懂第四題
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The art of doing mathematics consists in finding that special case which contains all the germs of generality
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| 2007-08-04 19:51 | |
