Cho S = 1/4+2/42+3/43+...+2014/42014
Chứng minh S < 1/2
K
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24 tháng 4 2022
4S=1+24+342+....+2014420134S=1+24+342+....+201442013
4S−S=3S=1+24+342+....+201442013−(14+242+343+....+201442014)4S−S=3S=1+24+342+....+201442013−(14+242+343+....+201442014)
3S=1+(24−14)+(342−242)+......+(201442013−201342013)−2014420143S=1+(24−14)+(342−242)+......+(201442013−201342013)−201442014
3S=1+14+142+143+.....+142013−2014420143S=1+14+142+143+.....+142013−201442014
đặt A=1+14+142+143+....+142023A=1+14+142+143+....+142023
4A−A=4+1+14+142+.....+142022−(1+14+142+....+142023)4A−A=4+1+14+142+.....+142022−(1+14+142+....+142023)
3A=4−1420233A=4−142023
A=43−13.42023A=43−13.42023
⇒3S=43−13.42023−201442024⇒3S=43−13.42023−201442024
⇒S=49−19.42023−20143.42024⇒S=49−19.42023−20143.42024
do 49<48=1249<48=12
⇒S=49−19.42023−20143.42024<48=12(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
15 tháng 7 2022
g: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{19}{20}=\dfrac{1}{20}\)
h: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot..\cdot\dfrac{100}{99}=\dfrac{100}{2}=50\)
f: \(A=1+\dfrac{1}{2^{2014}}\)
\(B=\dfrac{2^{2014}+1+1}{2^{2014}+1}=1+\dfrac{1}{2^{2014}+1}\)
mà \(2^{2014}< 2^{2014}+1\)
nên A>B
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 \(A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\)
=> \(4A=4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{2012}}\)
=> \(3A=4A-A=\left(4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{2012}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\right)\)
=> \(3A=4-\dfrac{1}{4^{2013}}< 4\)
=> \(A< \dfrac{4}{3}\)
=> \(3S< \dfrac{4}{3}\)
=> \(S< \dfrac{4}{9}< \dfrac{1}{2}\)
11 tháng 3 2022
\(4S=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2014}{4^{2013}}\)
\(4S-S=3S=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2014}{4^{2013}}-\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+....+\frac{2014}{4^{2014}}\right)\)
\(3S=1+\left(\frac{2}{4}-\frac{1}{4}\right)+\left(\frac{3}{4^2}-\frac{2}{4^2}\right)+......+\left(\frac{2014}{4^{2013}}-\frac{2013}{4^{2013}}\right)-\frac{2014}{4^{2014}}\)
\(3S=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+.....+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)
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 \(\frac{4}{9}< \frac{4}{8}=\frac{1}{2}\)
\(\Rightarrow S=\frac{4}{9}-\frac{1}{9.4^{2023}}-\frac{2014}{3.4^{2024}}< \frac{4}{8}=\frac{1}{2}\left(đpcm\right)\)
NP
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