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NV
1
22 tháng 3 2021
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
NV
Cho A = 1/2 + 1/3 + 1/4 + ... + 1/2017 B = 1/2015 + 2/2014 +3/2013 + ...+ 2015/2 + 2016/1 Tính B : A
1
22 tháng 3 2021
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
NN
1
12 tháng 2 2016
1/1+2+1/1+2+3+1/1+2+3+4+...+1/1+2+3+4+..+2015+1/50
=1/(2+1).2:2+1/(3+1).3:2+1/(4+1).4:2+...+1/(2015+1).2015:2+1/50
=2/2.3+2/3.4+2/4.5+..+2/2015.2016+1/50
=2(1/2.3+1/3.4+1/4.5+..+1/2015.2016)+1/50
=2(1/2-1/3+1/3-1/4+1/4-1/5+..+1/2015-1/2016)+1/50
=2(1/2-1/2016)+1/50
=1007/1008+1/50
=25679/25200
PL
2
20 tháng 9 2018
ta có: \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{2015^2}< \frac{1}{2014.2015};\frac{1}{2016^2}< \frac{1}{2015.1026};\frac{1}{2017^2}< \frac{1}{2016.2017}\)
=> 1/22 + 1/32 + 1/42 + ... + 1/20152 + 1/20162 + 1/20172 < 1/22 + (1/2.3 + 1/3.4 + ....+1/2014.2015 + 1/2015.2016 + 1/2016.2017)
= 1/4 + 1/2 - 1/2017 = 3/4- 1/2017 < 3/4
=> đ p c m
20 tháng 9 2018
ta có: \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{2015^2}< \frac{1}{2014.2015};\frac{1}{2016^2}< \frac{1}{2015.1026};\frac{1}{2017^2}< \frac{1}{2016.2017}\)
=> 1/22 + 1/32 + 1/42 + ... + 1/20152 + 1/20162 + 1/20172 < 1/22 + (1/2.3 + 1/3.4 + ....+1/2014.2015 + 1/2015.2016 + 1/2016.2017)
= 1/4 + 1/2 - 1/2017 = 3/4- 1/2017 < 3/4
=> đ p c m
LT
0
T
1
XO
8 tháng 9 2020
\(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2015^2}=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}\right)\)
\(=1-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{2015.2015}\right)>1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(=1-\left(1-\frac{1}{2015}\right)=1-\frac{2014}{2015}=\frac{1}{2015}\)
=> \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2015^2}>\frac{1}{2015}\left(\text{đpcm}\right)\)
TD
1
25 tháng 3 2016
1*2)^ -1+(2*3)^-1 + (3*4)^-1 +...+(2014*2015)^-1
1*2)^ -1+(2*3)^-1 + (3*4)^-1 +...+(2014*2015)^-1
1*2)^ -1+(2*3)^-1 + (3*4)^-1 +...+(2014*2015)^-1
ai tích mình mình tích lại
19 tháng 3 2017
A = \(\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{2013.2014.2015.2016}\)
3A =\(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{2013.2014.2015.2016}\)
3A = \(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{2013.2014.2015}-\dfrac{1}{2014.2015.2016}\)
3A = \(\dfrac{1}{1.2.3}-\dfrac{1}{2014.2015.2016}\)
3A = \(\dfrac{2014.2015.2016-6}{6.2014.2015.2016}\)
A=\(\dfrac{2014.2015.2016-6}{6.2014.2015.2016}:3\)
A=\(\dfrac{2014.2015.2016-6}{6.2014.2015.2016}.\dfrac{1}{3}\)
A=\(\dfrac{2014.2015.2016-6}{9.2014.2015.2016}\)
Mình không muốn rút gọn hơn vì nó sẽ quá cồng kềnh nên mình để tạm thế này nhé!