Question

2003/1*2+2003/2*3+2003/3*4+...+2003/2002*2003

Answer 1

Đặt A = 2003/1.2 + 2003/2.3 + 2003/3.4 + ... + 2003/2002.2003

A = 2003 . ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/2002.2003 )

A = 2003 . ( 1 - 1/2003 )

A = 2003 . 2002/2003

A = 2002

Answer 2

Đặt A = 2003/1.2 + 2003/2.3 + 2003/3.4 + ... + 2003/2002.2003

A = 2003 . ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/2002.2003 )

A = 2003 . ( 1 - 1/2003 )

A = 2003 . 2002/2003

A = 2002

Answer 3

bài này khó đấy

Answer 4

Ta có: A=2003/1.2+2013/2.3+2003/3.4+..+2003/2002.2003

=>      A=2003.(1/1.2+1/2.3+1/3.4+...+1/2002.2003)

=>      A=2003.(1-1/2+1/2-1/3+1/3-1/4+...+1/2002-1/2003)

=>      A=2003.(1-1/2003)

=>      A=2003.2002/2003

=>      A=2002

Vậy tổng A=2002

Answer 5

Đặt A = 2003/1.2 + 2003/2.3 + 2003/3.4 + ... + 2003/2002.2003

A = 2003 . ) 1/1.2 + 1/2.3 + 1/3.4 +... + 1/2002.2003

A = 2003 . ( 1 - 1/2003 )

A = 2003 . 2002/2003

A = 2002

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Answer 6

Đặt A = 2003/1.2 + 2003/2.3 + 2003/3.4 + ... + 2003/2002.2003

A = 2003 . ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/2002.2003 )

A = 2003 . ( 1 - 1/2003 )

A = 2003 . 2002/2003

A = 2002

Answer 7

=2002

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Answer 8

Đặt A = \(\frac{2003}{1}x2+\frac{2003}{2}x3+\frac{2003}{3}x4+...+\frac{2003}{2002}x2003\)
 

\(A=2003x\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{2002x2003}\right)\)

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

\(A=2003x\)\(\frac{2002}{2003}\)

\(A=2002\)