1/12+1/2.3+1/3,4+......+1/1999.2000
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
26 tháng 9 2020
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1999.2000}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{1999}+\frac{1}{1999}-\frac{1}{2000}\)
\(=\frac{1}{1}-\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{1999}+\frac{1}{1999}\right)-\frac{1}{2000}\)
\(=\frac{1}{1}+0+0+...+0-\frac{1}{2000}\)
\(=\frac{1}{1}-\frac{1}{2000}\)
\(=\frac{2000}{2000}-\frac{1}{2000}\)
\(=\frac{1999}{2000}\)
24 tháng 6 2015
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
TT
0
LT
5
31 tháng 3 2019
\(=\frac{1}{2}-\frac{1}{2000}=\frac{999}{2000}\)
Dạng tổng quát :
\(\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{n\left(n+1\right)}=\frac{1}{2}-\frac{1}{n+1}=\frac{n+1-2}{2\left(n+1\right)}=\frac{n-1}{2\left(n+1\right)}\)
31 tháng 3 2019
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{1999.2000}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{1999}-\frac{1}{2000}\)
\(=\frac{1}{2}-\frac{1}{2000}\)
\(=\frac{499}{2000}\)
NT
4
XO
3 tháng 11 2019
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1999.2000}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{1999}-\frac{1}{2000}\)
\(=1-\frac{1}{2000}\)
\(=\frac{1999}{2000}\)
Q
2
8 tháng 9 2016
Đặt A = 1.2+2.3 +3.4 +...+1999.2000
=> 3A = 1.2.(3 - 0) + 2.3.(4 - 1) + ..... + 1999.2000.(2001 - 1998)
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 1999.2000.2001
=> 3A = 1999.2000.2001
=> A = 1999.2000.2001 / 3
=> A = 2 666 666 000
23 tháng 9 2023
`#3107`
`a)`
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{1999\cdot2000}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\)
\(=1-\dfrac{1}{2000}\)
\(=\dfrac{1999}{2000}\)
`b)`
\(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{100\cdot103}?\)
\(=\dfrac{1}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{100\cdot103}\right)\)
\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{102}{103}\)
\(=\dfrac{34}{103}\)
`c)`
\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-....-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)
\(=\dfrac{8}{9}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)
\(=\dfrac{8}{9}-\dfrac{8}{9}\\ =0\)
VN
Võ Ngọc Phương
VIP
23 tháng 9 2023
b) Sửa đề:
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}.\left(\dfrac{103}{103}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}.\dfrac{102}{103}\)
\(=\dfrac{34}{103}\)
8 tháng 4 2019
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{1999.2000}\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{1999}+\frac{1}{2000}\)
\(S=1-\frac{1}{2000}\)
\(S=\frac{1999}{2000}\)
8 tháng 4 2019
Đây là bài làm của mk :
S = 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/1999 * 2000
=> S = 1 - 1/2 + 1/2 - 1/3 + ... + 1/1999 - 1/2000
=> S = 1 - 1 / 2000
=> S = 2000/2000 - 1/2000 = 1999/2000
Chúc bn học tốt !
HM
1
3 tháng 11 2019
=1/1-1/2+1/2-1/3+1/3-1/4+.........+1/1999-1/2000
=1/1-1/2000
=1999/2000<3/4
=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{1999\cdot2000}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...-\frac{1}{999}+\frac{1}{999}-\frac{1}{2000}\)
=\(\frac{1}{1}-\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{999}+\frac{1}{999}\right)-\frac{1}{2000}\)
=\(\frac{1}{1}+0+0+...+0-\frac{1}{2000}\)
=\(\frac{1}{1}-\frac{1}{2000}\)
=\(\frac{2000}{2000}-\frac{1}{2000}\)
=\(\frac{1999}{2000}\)