1+(-2)+3+(-4).+2001+(-2002)

1+(-2)+3+(-4).+2001+(-2002)

The final result is - 1001
It can be seen that there are a total of 2002 numbers, and the sum of every two values is equal to - 1;
2002 numbers can be divided into 1001 groups, and the sum of each group is - 1. So
The final result is - 1001

（1-2）（2-3）（3-4）… （2001-2002）=______ ．

The original formula = - 1 × (- 1) × (- 1) × × (- 1) = (- 1) 2001 = - 1, so the answer is: - 1

1-2+3-4+…… +2001-2002,

-1*2002/2=-1001

1 + 2 + 1 = 4 1 + 2 + 3 + 2 + 1 = 9 1 + 2 + 3 + 4 + 3 + 2 + 1 = 16 calculate 1 + 2 + 3 +... 2001 + 2002 + 2003 + 2002 + 2001 +... + 3 + 2 + 1

2003X2003===4012009

（1-2）（2-3）（3-4）… （2001-2002）=______ ．

The original formula = - 1 × (- 1) × (- 1) × × (- 1) = (- 1) 2001 = - 1, so the answer is: - 1

-What is 1 + 2-3 + 4-5 + 6.2001 + 2002

-1+2-3+4-5+6.-2001+2002
=(-1+2)+(-3+4)+(-5+6)+.+(-2001+2002)
=1+1+1+...+1
=1*1001
=1001

1-2+3-4+5-6+… +What is the value of 2001-2002___ ．

1-2+3-4+5-6+… +2001-2002, = (- 1) ×, = - 1 × 1001, = - 1001; so the answer is: - 1001

(1-1/2001)+(1-2/2001)+(1-3/2001)+······+(1-1999/2001)+(1-2000/2001)

(1-1/2001)+(1-2/2001)+(1-3/2001)+······+(1-1999/2001)+(1-2000/2001)
=2000-1/2000×（1+2+3+…… +1999+2000）
=2000-1/2000×（1+2000）×2000/2
=2000-1000.5
=999.5
That's all

2000*(199919991999/200020002000)+(2000*1999—2001*1998)/(2000^2—2001*1999)

2000*(199919991999/200020002000)+(2000*1999—2001*1998)/(2000^2—2001*1999)
=2000*[1999/2000]+[2000*1999-2000*1998-1998]/[2000^2-(2000^2-1)
=1999+[2000-1998]/1
=1999+2
=2001

2007 / 2006 compared with 2008 and 2007

1-2006/2007=1/2007
1-2007/2008=1/2008
1/2007>1/2008
2006 / 2007