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1 tháng 5 2017

\(S=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+....+\frac{2014}{4^{2014}}\)

\(4S=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2014}{4^{2013}}\)

\(4S-S=\left(1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2014}{4^{2013}}\right)-\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2014}{4^{2014}}\right)\)

\(3S=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)

\(12S=4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2012}}-\frac{2014}{4^{2013}}\)

\(12S-3S=\left(4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2012}}-\frac{2014}{4^{2013}}\right)-\left(1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\right)\)

\(9S=4-\frac{2014}{4^{2013}}-\frac{1}{4^{2013}}+\frac{2014}{4^{2014}}\)

\(9S=4-\frac{4028}{4^{2014}}-\frac{4}{4^{2014}}+\frac{2014}{4^{2014}}\)

\(9S=4-\frac{2010}{4^{2014}}< 4\)

\(\Rightarrow9S< 4\)

\(\Rightarrow S< \frac{4}{9}< 1\)(đpcm)

1 tháng 5 2017

Ta có :

\(S=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2014}{4^{2014}}\)( 1 )

\(4S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2014}{4^{2013}}\)( 2 )

Lấy ( 2 ) - ( 1 ) ta được :

\(3S=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)

gọi     \(B=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}\)( 3 )

\(4B=4+1+\frac{1}{4}+...+\frac{1}{4^{2012}}\)  ( 4 )

Lấy ( 4 ) - ( 3 ) ta được :

\(3B=4-\frac{1}{4^{2013}}\)

\(\Rightarrow B=\frac{4-\frac{1}{4^{2013}}}{3}=\frac{4}{3}-\frac{1}{4^{2013}.3}\)

\(\Rightarrow3S=\frac{4}{3}-\frac{1}{4^{2013}.3}-\frac{2014}{4^{2014}}\)

\(\Rightarrow S=\frac{\frac{4}{3}-\frac{1}{4^{2013}.3}-\frac{2014}{4^{2014}}}{3}=\frac{4}{9}-\frac{1}{4^{2013}.9}-\frac{2014}{4^{2014}.3}< \frac{4}{9}< 1\)

vậy \(S< 1\)

11 tháng 3 2022

Xét \(4S=1+\dfrac{2}{4}+\dfrac{3}{4^2}+\dfrac{4}{4^3}+...+\dfrac{2014}{4^{2013}}\)

=> \(3S=4S-S=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2014}{4^{2013}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+...+\dfrac{2014}{4^{2014}}\right)\)

=> \(3S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}-\dfrac{2014}{4^{2014}}< 1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\)

Đặt \(A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\)

=> \(4A=4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{2012}}\)

=> \(3A=4A-A=\left(4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{2012}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\right)\)

=> \(3A=4-\dfrac{1}{4^{2013}}< 4\)

=> \(A< \dfrac{4}{3}\)

=> \(3S< \dfrac{4}{3}\)

=> \(S< \dfrac{4}{9}< \dfrac{1}{2}\)

\(4S=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2014}{4^{2013}}\)

\(4S-S=3S=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2014}{4^{2013}}-\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+....+\frac{2014}{4^{2014}}\right)\)

\(3S=1+\left(\frac{2}{4}-\frac{1}{4}\right)+\left(\frac{3}{4^2}-\frac{2}{4^2}\right)+......+\left(\frac{2014}{4^{2013}}-\frac{2013}{4^{2013}}\right)-\frac{2014}{4^{2014}}\)

\(3S=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+.....+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)

đặt \(A=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{2023}}\)

\(4A-A=4+1+\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{2022}}-\left(1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2023}}\right)\)

\(3A=4-\frac{1}{4^{2023}}\)

\(A=\frac{4}{3}-\frac{1}{3.4^{2023}}\)

\(\Rightarrow3S=\frac{4}{3}-\frac{1}{3.4^{2023}}-\frac{2014}{4^{2024}}\)

\(\Rightarrow S=\frac{4}{9}-\frac{1}{9.4^{2023}}-\frac{2014}{3.4^{2024}}\)

do \(\frac{4}{9}< \frac{4}{8}=\frac{1}{2}\)

\(\Rightarrow S=\frac{4}{9}-\frac{1}{9.4^{2023}}-\frac{2014}{3.4^{2024}}< \frac{4}{8}=\frac{1}{2}\left(đpcm\right)\)

18 tháng 4 2016

=>4.S=1+2/4 +3/42+....+2014/42013

=>3.S=1+1/4+1/42+...+1/42013-2014/42014

=>12.S=4+1+1/4+......+1/42012-2014/42013

=>9.S=4-2014/42013-1/42013+2014/42014

=>9.S=4-(2015/42013-2014/42014) mà 2015/42013-2014/42014>0

=>9.S<4

=>S<4/9

=S<4/8

=>S<1/2

=>S<0,5

Vậy S<0,5   (ĐPCM)

23 tháng 1 2018

^ là dấu phân số nhé

cho A=1^1.2+1^2.3+...+1^2014.2015

1^1.2>1^4; 1^2.3>2^42; 1^3.4>3^43;...;1^2014.2015>2014^42014

mà A=1^1.2+1^2.3+...+1^2104.2015=1-1^2+1^2-1^3+1^3+...+1^2014-1^2015

A=1-1^2015=2014^2015

mà 2014^2015>1^2>S nên 1^2>S