Chứng minh: \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...-\frac{2014}{3^{2014}}<\frac{1}{5}\)
Những câu hỏi liên quan
Chứng minh:
\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...-\frac{2014}{3^{2014}}<\frac{1}{5}\)
Chứng minh: \(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...-\frac{2014}{3^{2014}}<\frac{1}{5}\)
Ta có : A = 1/3 - 2/3^2 + 3/3^3 - 4/3^4 +...- 2014/3^2014
=> 3A = 1 - 2/3 + 3/3^2 - 4/3^3 +...- 2014/3^2013
=> 4A = 1- 1/3 + 1/3^2 -...- 1/3^2013 - 2014/3^2014
Xét B = 1-1/3+1/3^2 -...- 1/3^2013
=> 3B = 3 - 1 + 1/3 -...- 1/3^2012
=> 4B = 3- 1/3^2013
=> B = (3- 1/3^2013)/4 < 3/4
=> 4A < 3/4 - 2014/3^2014< 3/4
=> A < 3/16 < 3/15 =1/5
Vậy A < 1/5 (đpcm)
Chúc bạn học tốt
Đúng 0
Bình luận (0)
Chứng minh: \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^2}-\frac{4}{3^4}+\)\(...-\frac{2014}{3^{2014}}<\frac{1}{5}\).
Cho tổng gồm 2014 số hạng, \(S=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2014}{4^{2014}}\). Chứng minh rằng \(S< \frac{1}{2}\).
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 \(A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\)
=> \(4A=4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{2012}}\)
=> \(3A=4A-A=\left(4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{2012}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\right)\)
=> \(3A=4-\dfrac{1}{4^{2013}}< 4\)
=> \(A< \dfrac{4}{3}\)
=> \(3S< \dfrac{4}{3}\)
=> \(S< \dfrac{4}{9}< \dfrac{1}{2}\)
Đúng 2
Bình luận (0)
\(4S=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2014}{4^{2013}}\)
\(4S-S=3S=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2014}{4^{2013}}-\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+....+\frac{2014}{4^{2014}}\right)\)
\(3S=1+\left(\frac{2}{4}-\frac{1}{4}\right)+\left(\frac{3}{4^2}-\frac{2}{4^2}\right)+......+\left(\frac{2014}{4^{2013}}-\frac{2013}{4^{2013}}\right)-\frac{2014}{4^{2014}}\)
\(3S=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+.....+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)
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 \(\frac{4}{9}< \frac{4}{8}=\frac{1}{2}\)
\(\Rightarrow S=\frac{4}{9}-\frac{1}{9.4^{2023}}-\frac{2014}{3.4^{2024}}< \frac{4}{8}=\frac{1}{2}\left(đpcm\right)\)
cho S=\(\frac{1}{4^1}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2014}{4^{2014}}\)
chứng minh S<\(\frac{1}{2}\)
^ 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
Đúng 0
Bình luận (0)
\(s=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+........+\frac{2014}{4^{2014}}.\)
CHỨNG MINH RẰNG : S < \(\frac{1}{2}\)
Chứng minh : \(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{2014^3}< \frac{1}{4}\).
Cho tổng gồm 2014 số hạng: \(S=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2014}{4^{2014}}\)
Chứng minh rằng S<0,5.
=>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)
Đúng 0
Bình luận (0)
cho tổng băngf 2014 số hạng :
S=\(\frac{1}{4}+\frac{2}{^{4^2}}+\frac{3}{^{4^3}}+...+\frac{2014}{4^{2014}}\)
CHỨNG MINH : S<\(\frac{1}{2}\)