Chứng minh rằng \(\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+...+\frac{1}{2012.1342}\)<1.5
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
11 tháng 5 2017
\(A=\dfrac{3}{1.4}+\dfrac{3}{2.6}+\dfrac{3}{3.8}+...............+\dfrac{1}{2012.1342}\)
\(A=\dfrac{3}{1.4}+\dfrac{3}{2.6}+\dfrac{3}{3.8}+...........................+\dfrac{3}{2012.4026}\)
\(A=\dfrac{6}{2.4}+\dfrac{6}{4.6}+\dfrac{6}{6.8}+..........................+\dfrac{6}{4024.4026}\)
\(A=3\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...................+\dfrac{2}{4024.4026}\right)\)
\(A=3\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+....................+\dfrac{1}{4024}-\dfrac{1}{4026}\right)\)
\(A=3\left(\dfrac{1}{2}-\dfrac{1}{4026}\right)\)
\(A=3.\dfrac{1}{2}-3.\dfrac{1}{4026}\)
\(A=1,5-\dfrac{3}{4026}< 1,5\)
DT
Đỗ Thanh Hải
CTVVIP
11 tháng 5 2017
Ta có
A = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{2.6}\) + \(\dfrac{3}{3.8}\) + ... + \(\dfrac{1}{2012.1342}\)
A = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{2.6}\) + \(\dfrac{3}{3.8}\) + ... + \(\dfrac{3}{2012.4026}\)
A = \(\dfrac{6}{2.4}\) + \(\dfrac{6}{4.6}\) + \(\dfrac{6}{6.8}\) + ... + \(\dfrac{6}{4024.4026}\)
A = \(3\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{4024.4026}\right)\)
A = \(3\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{4024}-\dfrac{1}{4026}\right)\)
A = \(3\left(\dfrac{1}{2}-\dfrac{1}{4026}\right)\)
A = 3.\(\dfrac{1}{2}\) - 3.\(\dfrac{1}{4026}\)
A = 1,5 - \(3.\dfrac{1}{4026}\) < 1,5
=> A < 1,5
=> đpcm
11 tháng 5 2017
\(A=\)\(\frac{3}{1.4}\)\(+\)\(\frac{3}{2.6}\)\(+\)\(\frac{3}{2.8}\)\(+\).........\(+\)\(\frac{1}{2012.1342}\)\(< 1,5\)
\(=\)\(\frac{3}{1.4}\)\(+\)\(\frac{3}{2.6}\)\(+\)\(\frac{3}{3.8}\)\(+\)............\(+\)\(\frac{3}{2012.4026}\)
\(=\)\(\frac{6}{2.4}\)\(+\)\(\frac{6}{4.6}\)\(+\)\(\frac{6}{6.8}\)\(+\)..............\(+\)\(\frac{6}{4024.4026}\)
\(=\)\(3.\)\(\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...........+\frac{2}{4024.4026}\right)\)
\(=\)\(3.\)\(\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{4024}-\frac{1}{4026}\right)\)
\(=\)\(3.\)\(\left(\frac{1}{2}-\frac{1}{4026}\right)\)
\(=\)\(3.\)\(\frac{1}{2}\)\(-\)\(3.\)\(\frac{1}{4026}\)
\(=\)\(1,5\)\(-\)\(\frac{3}{4026}\)\(< \)\(1,5\)
Vậy \(A< 1,5\)
UK
0
CT
0
TL
1
MV
1 tháng 5 2017
\(A=\dfrac{3}{1\cdot4}+\dfrac{3}{2\cdot6}+\dfrac{3}{3\cdot8}+...+\dfrac{1}{2012\cdot1342}\\ =\dfrac{3}{1\cdot4}+\dfrac{3}{2\cdot6}+\dfrac{3}{3\cdot8}+...+\dfrac{3}{2012\cdot4026}\\ =\dfrac{6}{2\cdot4}+\dfrac{6}{4\cdot6}+\dfrac{6}{6\cdot8}+...+\dfrac{6}{4024\cdot4026}\\ =3\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{4024\cdot4026}\right)\\ =3\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{4024}-\dfrac{1}{4026}\right)\\ =3\cdot\left(\dfrac{1}{2}-\dfrac{1}{4026}\right)\\ =3\cdot\dfrac{1}{2}-3\cdot\dfrac{1}{4026}\\ =1,5-\dfrac{3}{4026}< 1,5\)
Vậy \(A< 1,5\left(đpcm\right)\)
TM
0
H
0
\(\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+...+\frac{1}{2012.1342}\)
\(=\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+...+\frac{3}{2012.4026}\)
\(=\frac{6}{2.4}+\frac{6}{4.6}+\frac{6}{4.8}+...+\frac{6}{4024.4026}\)
\(=3\cdot\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{4024.4026}\right)\)
\(=3\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{4024}-\frac{1}{4026}\right)\)
\(=3\cdot\left(\frac{1}{2}-\frac{1}{4026}\right)\)
\(=3\cdot\frac{1}{2}-3\cdot\frac{1}{4026}\)
\(=1,5-\frac{3}{4026}< 1,5\)