Chứng minh rằng : \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{49x50}<1\)
Làm nhanh mình lick cho !
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
N
24 tháng 8 2017
\(\frac{1}{1x2}+\frac{1}{3x4}+....+\frac{1}{49x50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)+\left(-\frac{1}{2}-\frac{1}{4}-.....-\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}......+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+......+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\left(đpcm\right)\)
NT
5
2 tháng 5 2017
=1/2-1/3+1/3-1/4+.....+1/49+1/50
=1/2-1/50
=25/50-1/50
=24/50
=12/25
2 tháng 5 2017
\(\frac{1}{2x3}\)+ \(\frac{1}{3x4}\)+ ... + \(\frac{1}{49x50}\)
= \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ... + \(\frac{1}{49}\)- \(\frac{1}{50}\)
= \(\frac{1}{2}\)- \(\frac{1}{50}\)
=\(\frac{12}{25}\)
6 tháng 11 2017
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{8x9}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
=\(1-\frac{1}{9}\)
=\(\frac{8}{9}\)
OK XONG NHỚ CHO MIK NHA
LT
6 tháng 11 2017
\(\frac{1}{1\times2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+.......+\frac{1}{7x8}+\)\(\frac{1}{8x9}\)
=1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{8}-\frac{1}{9}\)
=1-\(\frac{1}{9}\)
=\(\frac{8}{9}\)
M
3
31 tháng 8 2020
\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(C=1-\frac{1}{2018}\)
\(C=\frac{2017}{2018}\)
31 tháng 8 2020
\(C=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+.....+\frac{1}{2017x2018}\)
Ta thấy \(\frac{1}{1x2}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{2x3}=\frac{1}{2}-\frac{1}{3}\)
.............................................
\(\frac{1}{2017x2018}=\frac{1}{2017}-\frac{1}{2018}\)
\(\Rightarrow C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2017}-\frac{1}{2018}\)
\(\Rightarrow C=\frac{1}{1}-\frac{1}{2018}\)
\(\Rightarrow C=\frac{2017}{2018}\)
Chúc bạn học tốt nhớ k mình nhá
BT
8
18 tháng 3 2016
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
k cho mình nha bạn
CT
3
H
29 tháng 4 2019
\(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(=1+\left(1-\frac{1}{2018}\right)\)
\(=1+\left(\frac{2018}{2018}-\frac{1}{2018}\right)\)
\(=1+\left(\frac{2017}{2018}\right)\)
\(=\frac{2018}{2018}+\frac{2017}{2018}=\frac{4035}{2018}\)
L
1 tháng 5 2019
\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}...+\frac{1}{2017\cdot2018}\)
\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(=1+\left(1-\frac{1}{2018}\right)\)
\(=1+\frac{2017}{2018}\)
\(=1+\frac{2017}{2018}\)
\(=\frac{4035}{2018}\)
1/1x2+1/2x3+...+1/49x50
=1-1/2+1/2-1/3+.....+1/49-1/50
=1-1/50(1)
Ta co 1(2)
So sanh (1) voi (2) ta thay 1-1/50<1
=>1/1x2+...+1/49x50<1
(Phuong phap khu)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
=\(\frac{1}{1}-\frac{1}{50}=\frac{50}{50}-\frac{1}{50}=\frac{49}{50}<1\)
Vậy \(\frac{49}{50}<1\)