ta có công thức 1.2+2.3+3.4+...+n.(n+1)=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

áp dụng công thức vào bài ta có: 1.2+2.3+3.4+...+2002.2003 = \(\frac{2002.2003.2004}{3}=2678684008\)

TL:

a)\(2+4+6+...+2000=\frac{\left(2+2000\right).\left[\left(2000-2\right):2+1\right]}{2}\) 

\(=1001000\)

Câu b tương tự nha bạn:)

c) Đặt 1.2+2.3+....+99.100 =A

\(3A=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)\) 

\(3A=1.2.3+2.3.4-1.2.3+...99.100.101-98.99.100\) 

\(3A=99.100.101\) 

\(A=333300\) 

Vậy .....

a) Đặt A= 2+4+6+...+1998+2000 

Ta có: A=(2+2000).1000:2

=> A=2002.1000:2

=> A=2002000:2

=> A=1001000

b) Đặt B= 5+9+13+...+1997+2001 

=> B=(2001+5).500:2

=> B=2006.500:2

=> B=1003000:2

=> B=501500

c)Đặt S= 1.2 + 2.3 + 3.4 + ...+ 99.100
=> 3S = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3
3S= 1.2.3+2.3(4-1)+3.4(5-2)+...+98.99(100-97)+99.100(101-98)
3S= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-97.98.99+99.100.101-98.99.100
3S = 99.100.101 => 3S = 3.33.100.101 
=> S=33.100.101= 333300

a) 2+4+6+...+1998+2000 

=(2+2000).1000:2

=2002.1000:2

=1001000

b) 5+9+13+...+1997+2001 

=(2001+5).500:2

=2006.500:2

=501500

c)đặt A=1.2 + 2.3 + 3.4 + ...+ 99.100
=> 3A = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3
3A= 1.2.3+2.3(4-1)+3.4(5-2)+...+98.99(100-97)+99.100(101-98)
3A= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-97.98.99+99.100.101-98.99.100
3A = 99.100.101 
=> A=33.100.101= 333300

\(M=1.2+2.3+3.4+...+2002.2003\)

\(3.M=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+2002.2003.\left(2004-2001\right)\)

\(3.M=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-...+2002.2003.2004-2001.2002.2003\)

\(3.M=2002.2003.2004\)

\(M=2002.2003.2004:3=2002.2003.668\)

\(M=2678684008\)

M = 1 . 2 + 2 . 3 + 3 . 4 + ... + 2002 . 2003

3M = 1 . 2 . 3 + 2 . 3 . 4 + 3 . 4 . 3 + ... + 2002 . 2003 . 3

3M = 1 . 2 ( 4  - 1 ) + 2 . 4 ( 5 - 2 ) + 3 . 4 ( 6 - 3 ) + ... + 2002 . 2003 ( 2005 - 2002 )

3M = 1 . 2 . 3 + 2 . 3 . 4 - 1 . 2 . 3 + 3 . 4 . 5 - 2 . 3 . 4 + .... - 2002 . 2003 . 2004 + 2004 . 2005 . 2006

3M = 2005 . 2006 . 2007

3M = 2005 . 2006 . 889 . 3

M = 2005 . 2006 . 889

M = 4022030

M = 1.2 + 2.3 + 3.4 + 4.5 + 5.6 + ..... + 2002.2003

3.M = 1.2.3 + 2.3.3 + 3.4.3 + 4.5.3 + 5.6.3 + ..... + 2002.2003.3

3.M = 1.2.3 +  2.3. ( 4 - 1 ) + 3.4. ( 5 - 2 ) + 4.5. ( 6 - 3 ) + 5.6. ( 7 - 4 ) + ......... + 2002.2003. ( 2004 - 2001 )

3.M = 1.2.3 + 2.3.4 -1.2.3 + 3.4.5 - 2.3.4 + 4.5.6 - 3.4.5 + 5.6.7 - 4.5.6 + ...... + 2002.2003.2004 - 2001.2002.2003

3.M = 2002.2003.2004

Vậy M = 2002.2003.2004 : 3

M = 2678684008

Dùng công thức \(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\) là ra nha bạn 
Tức \(\frac{2002.2003.2004}{3}=2678684008\)

(1-1/2).(1-1/3).(1-1/4)...(1-1/2002).x=1-1/1.2-1/2.3-1/3.4-...-1/2002.2003 ghi loi giai nha ae

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