解答*數學難題* ###十萬火急### 盡快回答 PLEASE

1. (1+n)n/2是1至n的和
(1+n)n/2>1991>[(1+n)n/2]-n
n+n^2>3982 > n^2
n^2+n-3982>0 and 3982>n^2
(n+63.6)(n-62.6) > 0 and n<63.1
n<63.6 or n>62.6 and n<63.1
n=63
1至63的和=(1+63)63/2
=2016
漏加的那個數=2016-1991=25

2. 1/1990大約=1/1991大約=1/1992......大約=1/1999
1/1990+1/1991+.....1/1999大約=10/199x
A=1/(10/199x)
=199x/10
=199.x
A 的整數部分是199

2007-03-31 01:00:25 補充：
Q1有少少野改1. (1+n)n/2是1至n的和(1+n)n/2>1991>[(1+n)n/2]-nn+n^2>3982>n^2-nn^2+n-3982>0 and n^2-n-3982>0(n+63.6)(n-62.6)>0 and (n-63.6)(n+62.6)>0n<-63.6 or n>62.6 and n<63.6 and n>-62.6n=631至63的和=(1+63)63/2=2016漏加的那個數=2016-1991=25

Q1 設小明按了 n 個數字，第 q 個數字打漏了，

(1 + 2 + ... + n) - q = 1991

__n(n+1)__ - q = 1991
..........2

n(n+1) - 2q = 3982

By method of bisection, 63X64 = 4032 and 62X63=3906

所以， n = 63

q = __4032 - 3982__ = 25
..................2


Q2
......______________1_______________
A= __1__ + __1__ + __1__ + ...... + __1__
.......1990......1991.......1992................1999

___1____ ..................___1____
...._10_ ........ < A <......_10_
....1999...........................1990

_1999_ < A < _1990_
.....10.....................10

199 _9_ < A < 199
.........10

所以 A 的整數部分為 199

Q1.
1至10的總和
=1+10除2再乘10
=55
11至20的總和
=11+20除2再乘10
=155
21至30的總和
=21+30除2再乘10
=255
如此類推﹕
加至63
=2016
而2016減1991
=25
所以答案是25