ban chi can nhan tat co cac do hang voi3 xong sau do ban tinh 

ban chi can nhan cac so hang voi 3 la duoc

đặt là S rồi nhân S và dãy với 3

ta có :

Đặt A = 1.2 + 2.3 + 3.4 + ..... + 1999.2000

=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ..... + 1999.2000.3

=> 3A = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + ...... + 1999.2000.( 2001 - 1998 )

=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ..... + 1999.2000.2001 - 1998.1999.2000

=> 3A = 1999.2000.2001

=> A = \(\frac{1999.2000.2001}{3}\)

ket qua A = \(\frac{1999.2000.2001}{3}\) ban nha

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1999.2000}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{1999}+\frac{1}{1999}-\frac{1}{2000}\)

\(=\frac{1}{1}-\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{1999}+\frac{1}{1999}\right)-\frac{1}{2000}\)

\(=\frac{1}{1}+0+0+...+0-\frac{1}{2000}\)

\(=\frac{1}{1}-\frac{1}{2000}\)

\(=\frac{2000}{2000}-\frac{1}{2000}\)

\(=\frac{1999}{2000}\)

1/1.2 + 1/2.3 + 1/3.4 + ... + 1/1999.2000

= 1 -1/2+1/2-1/3+1/3-1/4+....+1/1999-1/2000

= 1- 1/2000

= 1999/2000

(1999.2000.2001)/3

Đúng đấy !

giải:

đặt A=1.2+2.3+3.4+....+1999.2000

=>3A=1.2.3+2.3.3+3.4.3+...+1999.2000.3

=>3A=1.2.3+2.3.(4-1)+3.4.(5-2)+.....+1999.2000(2001-1998)

=>3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+1999.2000.2001-1998.1999.2000

=>3A=1999.2000.2001

=>A=\(\dfrac{1999.2000.2001}{3}\)

1.2 = 1/3. 1.2.3 - 0.1.2

2.3 = 1/3. 2.3.4 - 1.2.3

................................

1999.2000 = 1/3. 1999.2000.2001 - 1998.1999.2000 = 3998000

Quậy A = 3998000

Hãy k đúng cho mik nha!!!!!!!!!!

1.2+2.3+3.4+.........+1999.2000

=>3A=1.2.3+2.3.3+3.4.3+....+1999.2000.3

=>3A=1.2.3+2.3.(4-1)+3.4.(5-2)+.....+1999.2000.(2001-1998)

=>3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+1999.2000.2001-1998.1999.2000

=>3A=1999.2000.2001

=>A=7999998000:3

=>A=2666666000