4/1.3 +4/3.5 +......+4/2015.2017
K
Khách
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31 tháng 3 2017
M = \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{2015.2017}\)4/1.3 + 4/3.5 + 4/5.7 + ... + 4/2015.2017
M = \(2.\frac{2}{1.3}+2.\frac{2}{3.5}+2.\frac{2}{5.7}+...+2.\frac{2}{2015.2017}\) 2 . 2/1.3 + 2 . 2/3.5 + 2 . 2/5.7 + ... + 2 . 2/2015.2017
M = 2 . ( 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2015.2017 )
M = 2 . ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2015 - 1/2017 )
M = 2 . ( 1 - 1/2017 )
M = 2 . 2016/2017
M = 4032/2017
31 tháng 3 2017
\(M=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(M=2\left(1-\frac{1}{2017}\right)\)
\(M=\frac{4032}{2017}\)
PM
1
18 tháng 3 2016
A=4/3+9/8+16/15+..............+4064256/4064255
A=1+1/3+1+1/8+1/15+...............+1/4064255
A=(1+1+...+1)+(1/3+1/8+...+1/406255) (có 2015 số 1)
A=2015+(1/1.3+1/2.4+...........+1/2015.2017)
A=2015+1/2(1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7+....+1/2012-1/2014+1/2013-1/2015+1/2014-1/2016+1/2015-1/2017)
A=2015+1/2(1+1/2-1/2016-1/2017)
A=2015,749504
k cho mình nhé mình k lại cho
9 tháng 12 2019
Đặt A = \(\frac{1.3+2}{2^2}+\frac{2.4+2}{3^2}+\frac{3.5+2}{4^2}+...+\frac{2010.2012+2}{2011^2}+\frac{2015.2017+2}{2016^2}\)
\(=\frac{\left(2-1\right)\left(2+1\right)+2}{2^2}+\frac{\left(3-1\right)\left(3+1\right)}{3^2}+...+\frac{\left(2016-1\right)\left(2016+1\right)+2}{2016^2}\)
\(=\frac{2^2-1+2}{2^2}+\frac{3^2-1+2}{3^2}+....+\frac{2016^2-1+2}{2016^2}\)
\(=\frac{2^2+1}{2^2}+\frac{3^2+1}{3^2}+...+\frac{2016^2+1}{2016^2}\)
\(=\left(1+1+...+1\right)+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}\right)\)(2015 hạng tử 1)
\(=2015+\left(\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{2016.2016}\right)\)
\(< 2015+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\right)\)
\(=2015+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\right)=2015+\left(1-\frac{1}{2016}\right)\)
= 2015 + 1 + 1/2016
= 2016 + 1/2016 < 2017
=> A < 2017 (ĐPCM)
18 tháng 5 2018
\(\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{2015.2017}\)
\(=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2015.2017}\right)\)
\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=\frac{3}{2}.\left(1-\frac{1}{2017}\right)\)
\(=\frac{3}{2}.\frac{2016}{2017}\)
\(=\frac{3024}{2017}\)
_Chúc bạn học tốt_
22 tháng 4 2023
=1-1/3+1/3-1/5+...+1/2015-1/2017
=1-1/2017
=2016/2017
H
9 tháng 5 2016
B=1/1.3+1/3.5+...+1/2015.2017
B= 1/2 . 2. ( 1/1.3+1/3.5 + .... + 1/2015 .2017)
B = 1/2 . ( 2/1.3 + 2/3.5 + ......+ 2/2015.2017)
B = 1/2. ( 1/1+ -1/3 + 1/3 + -1/5 + 1/5 +....+ 1/2015 + -1/2017)
B= 1/2 . ( 1/1 + -1 / 2017) = 1/2 . 2016 / 2017 = 2016 / 4034
Vậy B = 2016 / 4034 nha bn.pham hong thai
\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+...........+\dfrac{4}{2015.2017}\)
\(=2\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+............+\dfrac{2}{2015.2017}\right)\)
\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+.........+\dfrac{1}{2015}-\dfrac{1}{2017}\right)\)
\(=2\left(1-\dfrac{1}{2017}\right)\)
\(=2.\dfrac{2016}{2017}=\dfrac{4032}{2017}\)
\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{2015.2017}\)
= 2.(\(\dfrac{2}{1.3}+\dfrac{1}{3.5}+...+\dfrac{2}{2015.2017}\))
= 2.(1 - \(\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\))
= 2.(1 - \(\dfrac{1}{2017}\))
= 2.\(\dfrac{2016}{2017}\)
= \(\dfrac{4032}{2017}\)
@Nguyễn Thị Ngọc Anh