奧數10點

First , rearrange the terms :

(1997 + 1995 + 1993 + ... + 3 + 1) - (2 + 4 + 6 + ... + 1994 + 1996)

= ( 1997 - 1996) + ( 1995 - 1994 ) + ( 1993 - 1992) + ... + ( 3 - 2) + 1



As there are 999 1s ,

so

(1997 + 1995 + 1993 + ... + 3 + 1) - (2 + 4 + 6 + ... + 1994 + 1996)

= ( 1997 - 1996) + ( 1995 - 1994 ) + ( 1993 - 1992) + ... + ( 3 - 2) + 1

= 999 x 1

= 999

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Just to add some more :
The above , you are wrong.
As there are (1997 - 1 )/ 2 + 1 = 998 terms in
(1997 + 1995 + 1993 + ... + 3 + 1)
and there are (1996 - 1 )/ 2 + 1 = 997 terms in
(2 + 4 + 6 + ... + 1994 + 1996)
Yourt answer is WRONG !

(1997 + 1995 + 1993 + ... + 3 + 1) - (2 + 4 + 6 + ... + 1994 + 1996)
=1997-1996+1995-1994.....+3-2+1
=1+1+1+1+1...+1+1+1
=999

neglect the last term,the no. of terms is 1996/2=998
therefore there are 998 terms in which each term is equal to 1
So the final answer is:
(1997 + 1995 + 1993 + ... + 3 + 1) - (2 + 4 + 6 + ... + 1994 + 1996)
=(1997 + 1995 + 1993 + ... + 3 + 1) - (1996 + 1994 + ...+ 6+ 4 +2 )
=(1997-1996)+(1995-1994)+(1993-1992)+...+(3-2)+1
=1x998+1
=999

原式
=1997+(1995-1996)+(1993-1994)+...+(3-4)+(1-2)
=1997-1-1-1-1-1-1(998個1)
=1997-998
=999

觀察些便行:

1996裹有998個雙數, (1997 + 1995 + 1993 + ... + 3 + 1)<這裡不要了1,變了(1997-1996), (1995-1994)如此一來, 減減減減減變得了數字的數目:998, 後將不要了的1補上, 便變了999

問題) 1997 + 1995 + 1993 + ... + 3 + 1) - (2 + 4 + 6 + ... + 1994 + 1996)=??

[題解: 這題需分別運用兩次等差數例和公式來計,等差數例和公找出「1997 + 1995 + 1993 + ... + 3 + 1」的值,再用多次等差數例和公式找出「2 + 4 + 6 + ... + 1994 + 1996」的值,將前值減後值即可得出答案!」

等差數例和公式: S(n) = n/2 [ 2a + (n-1)d] *(n:項數、a:首項、d:公差 )
項數公式: T(n) = a + (n-1) d *(若要找整數例有多少項,T(n)的值等於數例的最尾一項)

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回答:
(1997 + 1995 + 1993 + ... + 3 + 1) - (2 + 4 + 6 + ... + 1994 + 1996)

『先計(1997 + 1995 + 1993 + ... + 3 + 1) :
先找出n(項數)*[即由1997+1995+1993加到1有幾多個數]』

T(n) = a + (n-1)(d) = 尾項
T(n) = 1997 + (n-1)(1995 - 1997) = 1
1997 + (n-1)(-2) = 1
1997 - 2n + 2 = 1
1997 + 2 + 1 = 2n
2000 = 2n
n = 1000

*[將所有已計出的值代入S(n):n = 1000, a = 1997, d = -2]
S(1000) = n/2 [ 2a + (n-1)d]
= 1000 / 2[( 2(1997)+(1000-1)(-2)]
= 500 [3994 + 999(-2)]
= 500(3994 - 1998)
= 500(1996)
= 998000
∴(1997 + 1995 + 1993 + ... + 3 + 1)的值是 998000。


(2 + 4 + 6 + ... + 1994 + 1996)
找出項數:
T(n) = a + (n-1) (d) = 尾項
T(n) = 2 + (n-1)(4-2) = 1996
2+(n-1)(2) = 1996
2+ 2n -2 =1996
2n = 1996
n = 1996 / 2
n = 998

*[將所有已計出的值代入S(n): n = 998, a = 2, d = 2]
S(998) = 998 / 2 [ 2(2) + (998 - 1)(2)]
= 499[4 + 997(2)]
= 499 (4 + 1994)
= 499(1998)
= 997002
∴( 2 + 4 + 6 + ... + 1994 + 1996)的值是 997002。

∴(1997 + 1995 + 1993 + ... + 3 + 1) - (2 + 4 + 6 + ... + 1994 + 1996)
= 998000 - 997002
= 998

(1997 + 1995 + 1993 + ... + 3 + 1) - (2 + 4 + 6 + ... + 1994 + 1996)=999

998001-997002=999

(1997 + 1995 + 1993 + ... + 3 + 1) - (2 + 4 + 6 + ... + 1994 + 1996)=-997

(1997 + 1995 + 1993 + ... + 3 + 1) - (2 + 4 + 6 + ... + 1994 + 1996) = 1000
為甚麼答案是這樣？解釋如下：
首先，我們要看出這個規律，發現1 - 2 = -1、3 - 4 = -1……由於有(1995+1)/2次，即是998次，所以答案是-1 x 998 = -998。
但是，我們要注意的是還有1997尚未處理，所以只需將-997 + 1997 = 1000，便知道答案。

2007-04-19 20:40:42 補充：
更正的地方：-997 1997 = 1000– -998 1997 = 999（答案與下面那位朋友所答的相同，只不過是方法不同）

2007-04-19 20:43:13 補充：
更正的地方：-997加1997 = 1000 to -998加1997 = 999（答案與下面那位朋友所答的相同，只不過是方法不同）因加號不能顯示，所以只好用中文字代替。