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A = 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/2017.2019
A = 1/2 (1 - 1/3 + 1/3 - 1/5 + 1/5 - ... - 1/2019)
A = 1/2 (1 - 1/2019)
A = 1/2 . 2018/2019
A = 1009/2019
@Cỏ
\(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2017\cdot2019}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2019}\right)=\frac{1}{2}\cdot\frac{2018}{2019}\)
\(=\frac{1009}{2019}\)
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{35.37}=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{35}-\frac{1}{37}\)
\(=\frac{1}{1}-\frac{1}{37}<1\text{ Vậy }A<1\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(\Rightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(\Rightarrow2A=1-\frac{1}{101}\)
\(\Rightarrow2A=\frac{100}{101}\)
\(\Rightarrow A=\frac{100}{101}:2\)
\(\Rightarrow A=\frac{50}{101}.\)
Chúc bạn học tốt!
\(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\\ A=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right)\\ A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\\ A=\frac{1}{2}\left(1-\frac{1}{101}\right)\\ A=\frac{1}{2}\cdot\frac{100}{101}\\ A=\frac{50}{101}\)
\(B=1-2+3-4+...+49-50\\ B=\left(1-2\right)+\left(3-4\right)+...+\left(49-50\right)\\ B=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\text{ (có 25 số -1)}\\ B=\left(-1\right)\cdot25=-25\)
a) Đặt B= 1/1.3 + 1/3.5 + 1/5.7 + .....+ 1/19.21
Ta có: 2B= 2/1.3 + 2/3.5 + 2/5.7 + ....+ 2/19.21
= 1- 1/3 + 1/3-1/5 + 1/5-1/7 +....+ 1/19-1/21
= 1-1/21 = 20/21
=> B= 20/21 : 2 => B= 10/21
b) Như trên, ta có: 2A= 1- (1/2n + 1) => A=( 1-1/2n+1).1/2
=> A= 1/2- 1/2n+1
=> A< 1/2 ( đpcm )
Đặt \(A=\frac{1}{1.3}+\frac{1}{2.4}+...+\frac{1}{8.10}\)
\(2A=\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\)
\(2A=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\)
\(2A=1-\frac{1}{10}\)
\(2A=\frac{9}{10}\)
\(A=\frac{9}{10}:2=\frac{9}{20}\)
=\(\frac{1}{2}\left(\frac{2}{1.3}+...+\frac{2}{8.10}\right)\)
=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}...+\frac{1}{8}-\frac{1}{10}\right)\)
( chắc chắn có số trái dấu ở phía sau, nên còn lại như sau)
=\(\frac{1}{2}\left(1-\frac{1}{10}\right)=\frac{1}{2}.\frac{9}{10}=\frac{9}{20}\)
6Q = 1.3.6 + 3.5.(7-1) + 5.7.(9-3) + ... + 1999.2001.(2003-1997)
6Q = 18 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ... + 1999.2001.2003 - 1997.1999.2001
6Q = (18 + 3.5.7 + 5.7.9 + ... + 1999.2001.2003) - (1.3.5 + 3.5.7 + ... + 1997.1999.2001)
6Q = 18 + 1999.2001.2003 - 1.3.5
6Q = 18 + 1999.2001.2003 - 15
6Q = 3 + 8011997997
6Q = 8011998000
Q = 1335333000
\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}.\frac{2018}{2019}\)
\(=\frac{2018}{4038}\)
\(\Rightarrow\frac{2018}{4038}< \frac{1}{2}\)( lấy máy tính )
\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{2017.2019}\)
\(\Rightarrow M=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2019}\)
\(\Rightarrow M=1-\frac{1}{2019}\)
\(\Rightarrow M=\frac{2019}{2019}-\frac{1}{2019}\)
\(\Rightarrow M=\frac{2018}{2019}\)
Có \(\frac{2018}{2019}=\frac{2018.2}{2019.2}=\frac{4036}{4038}\)
\(\frac{1}{2}=\frac{1.2019}{2.2019}=\frac{2019}{4038}\)
Mà \(\frac{4036}{4038}< \frac{2019}{4038}\Rightarrow M< \frac{1}{2}\)
Vậy M < \(\frac{1}{2}\)
=\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\right)\)
= \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+....+\frac{1}{8}-\frac{1}{10}\right)\)
= \(\frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\right)\)
=\(\frac{29}{45}\)
A= 1/1.3+1/3.5+1/5.7+.....+1/2013.2015
2A=2/1.3+2/3.5+2/5.7+......+2/2013.2015
A=1/1-1/3+1/3-1/5+1/5-1/7+....+1/2013-1/2015
A=1-1/2015=2014/2015
=>A=2014/2015:2
=>A=2014/4030
I K MK MK K LI 3 K
\(A=\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{37\cdot39}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{37}-\dfrac{1}{39}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{38}{39}< \dfrac{1}{2}\)