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 = \(\frac{2002.2003.2004}{3}=2678684008\)

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

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

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

\(3.M=2002.2003.2004\)

\(M=2002.2003.2004:3=2002.2003.668\)

\(M=2678684008\)

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 \(\frac{2002.2003.2004}{3}=2678684008\)

\(\frac{2000}{1.2}+...+\frac{2000}{2002.2003}\)
\(=2000.\left(\frac{1}{1.2}+....+\frac{1}{2002.2003}\right)\)
\(=2000.\left(\frac{1}{1}-\frac{1}{2}+...+\frac{1}{2002}-\frac{1}{2003}\right) \)
\(=2000.\left(\frac{1}{1}-\frac{1}{2003}\right)=2000.\frac{2002}{2003}\)

đặt A=200/1.2+200/2.3+200/3.4+...+200/2002.2003

A:2000 = 1-1/2+1/2-1/3+...+1/2002-1/2003

A:2000=1-1/2003

A:2000=2002/2003

A=....

k nhe

(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

\(\frac{2003}{1\cdot2}+\frac{2003}{2\cdot3}+...+\frac{2003}{2002\cdot2003}\)

\(=2003\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2002\cdot2003}\right)\)

\(=2003\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{2002}-\frac{1}{2003}\right)\)

\(=2003\cdot\left(1-\frac{1}{2003}\right)\)

\(=2003\cdot\frac{2002}{2003}\)

\(=\frac{2003\cdot2002}{2003}\)

\(=2002\)