Bài 1:

\(2^{49}=\left(2^7\right)^7=128^7;5^{21}=\left(5^3\right)^7=125^7\\ Vì:128^7>125^7\Rightarrow2^{49}>5^{21}\)

Bài 2:

\(a,S=1+3+3^2+3^3+...+3^{99}\\ =\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)\\ =40+3^4.40+...+3^{96}.40\\ =40.\left(1+3^4+...+3^{96}\right)⋮40\\ b,S=1+4+4^2+4^3+...+4^{62}\\ =\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{60}.\left(1+4+4^2\right)\\ =21+4^3.21+...+4^{60}.21\\ =21.\left(1+4^3+...+4^{60}\right)⋮21\)

Bài 1 :

\(2^{49}=\left(2^7\right)^7=128^7\)

\(5^{21}=\left(5^3\right)^7=125^7\)

mà \(125^7< 128^7\)

\(\Rightarrow2^{49}>5^{21}\)

Bài 2 :

a) \(S=1+3+3^2+3^3+...3^{99}\)

\(\Rightarrow S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)...+3^{96}\left(1+3+3^2+3^3\right)\)

\(\Rightarrow S=40+40.3^4+...+40.3^{96}\)

\(\Rightarrow S=40\left(1+3^4+...+3^{96}\right)⋮40\)

\(\Rightarrow dpcm\)

b) \(S=1+4+4^2+4^3+...4^{62}\)

\(\Rightarrow S=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...4^{60}\left(1+4+4^2\right)\)

\(\Rightarrow S=21+4^3.21+...4^{60}.21\)

\(\Rightarrow S=21\left(1+4^3+...4^{60}\right)⋮21\)

\(\Rightarrow dpcm\)

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a) \(S=1+2+2^2+..+2^{2022}\)

\(2S=2+2^2+2^3+...+2^{2023}\)

\(2S-S=2+2^2+2^3+...+2^{2023}-1-2-2^2-...-2^{2022}\)

\(S=2^{2023}-1\)

b) \(S=3+3^2+3^3+...+3^{2022}\)

\(3S=3^2+3^3+...+3^{2023}\)

\(3S-S=3^2+3^3+....+3^{2023}-3-3^2-...-3^{2022}\)

\(2S=3^{2023}-3\)

\(\Rightarrow S=\dfrac{3^{2023}-3}{2}\)

c) \(S=4+4^2+4^3+...+4^{2022}\)

\(4S=4^2+4^3+...+4^{2023}\)

\(4S-S=4^2+4^3+...+4^{2023}-4-4^2-...-4^{2022}\)

\(3S=4^{2023}-4\)

\(S=\dfrac{4^{2023}-4}{3}\)

d) \(S=5+5^2+...+5^{2022}\)

\(5S=5^2+5^3+...+5^{2023}\)

\(5S-S=5^2+5^3+...+5^{2023}-5-5^2-...-5^{2022}\)

\(4S=5^{2023}-5\)

\(S=\dfrac{5^{2023}-5}{4}\)

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)

^ 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

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

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

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