1/1.2+1/2.3+1/3.4+......+1/2003.2004=1/1-1/2+1/2-1/3+1/3-1/4+......+1/2003-1/2004

                                                          =1/1-1/2004

                                                          =2003/2004

1/1.2+1/2.3+1/3.4+.......1/2003.2004

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\frac{1}{2004}\)

\(=\frac{2003}{2004}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\frac{1}{2004}\)

\(=\frac{2003}{2004}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\frac{1}{2004}\)

\(=\frac{2003}{2004}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2003\cdot2004}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{2003}+\frac{1}{2003}-\frac{1}{2004}\)

\(=1-\frac{1}{2004}\)

\(=\frac{2003}{2004}\)

\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ ...+ \(\frac{1}{2003.2004}\)

= 1 - \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+...+ \(\frac{1}{2003}\)- \(\frac{1}{2004}\)

= 1 - \(\frac{1}{2004}\)

= \(\frac{2003}{2004}\)

= 

ai tích mình mình tích lại cho

tinh tong A= 1.2^2+2.3^2+3.4^2+........+2018.2019^2

=> 3A = 3 [ 1.2 + 2.3 + 3.4 + ... + (n-1).n ]

=> 3A = 1.2.3 + 2.3.3 + 3.4.3 +... + 1001.1002.3 

=> 3A =  1.2.3 + 2.3 . ( 4-1 ) +3.4.( 5-2 ) + ... + 1001.1002 ( 1003-1000 )

=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +... + 1001.1002 .1003 - 1000.1001.1002

=> 3A = 1001.1002.1003 

=> A = 1001 . 1002 . 1003 : 3 

=> A = ?

cái nay có trong sách bạn ak

chữ số tận cùng là 8

các bạn ơi giúp nha năn nỉ đó

k mk đi

ai k mk

mk k lại

thanks

a. Số số hạng: ( 601 - 1 ) : 3 + 1 = 201

Tổng :

1 + 4 + 7 + .... + 601

= ( 1 + 601 ) x 201 : 2 

= 602 x 201 : 2

= 60501

\(P=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(P=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(P=\frac{1}{1}-\frac{1}{100}\)

\(P=\frac{99}{100}\)

\(HT\)

\(P=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{99.100}\)

\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(P=1+\left(\dfrac{-1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{-1}{3}+\dfrac{1}{3}\right)+..+\left(\dfrac{-1}{99}+\dfrac{1}{99}\right)+\dfrac{-1}{100}\)

\(P=1+0+0+....+0+\dfrac{-1}{100}\)

\(P=1+\dfrac{-1}{100}\)

\(P=\dfrac{99}{100}\)

cảm ơn nha