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)

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

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

=>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)

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