\(B=1+2+2^2+\cdots+2^{2017}\)

=>\(2B=2+2^2+2^3+\cdots+2^{2018}\)

=>\(2B-B=2+2^2+\cdots+2^{2018}-1-2-\cdots-2^{2017}\)

=>\(B=2^{2018}-1\)

\(\frac{A}{B}=\frac{2^{2022}+2^{2021}}{2^{2018}-1}\)

\(=\frac{2^{2022}-2^4+2^{2021}-2^3+2^4+2^3}{2^{2018}-1}=2^4+2^3+\frac{2^4+2^3}{2^{2018}-1}\)

=>Dư là \(2^4+2^3=16+8=24\)

Sửa đề: A=2+2^2+2^3+...+2^2017

=>2*A=2^2+2^3+2^4+...+2^2018

=>2A-A=2^2018-2

=>A=2^2018-2

a.

$S=1+2+2^2+2^3+...+2^{2017}$
$2S=2+2^2+2^3+2^4+...+2^{2018}$

$\Rightarrow 2S-S=(2+2^2+2^3+2^4+...+2^{2018}) - (1+2+2^2+2^3+...+2^{2017})$

$\Rightarrow S=2^{2018}-1$

b.

$S=3+3^2+3^3+...+3^{2017}$
$3S=3^2+3^3+3^4+...+3^{2018}$

$\Rightarrow 3S-S=(3^2+3^3+3^4+...+3^{2018})-(3+3^2+3^3+...+3^{2017})$

$\Rightarrow 2S=3^{2018}-3$
$\Rightarrow S=\frac{3^{2018}-3}{2}$
 

Câu c, d bạn làm tương tự a,b. 

c. Nhân S với 4. Kết quả: $S=\frac{4^{2018}-4}{3}$

d. Nhân S với 5. Kết quả: $S=\frac{5^{2018}-5}{4}$

Ta có: \(\frac{1}{21}>\frac{1}{40};\frac{1}{22}>\frac{1}{40};\ldots;\frac{1}{39}>\frac{1}{40};\frac{1}{40}=\frac{1}{40}\)

Do đó: \(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+\cdots+\frac{1}{40}=\frac{20}{40}=\frac12\) (1)

Ta có: \(\frac{1}{41}>\frac{1}{80};\frac{1}{42}>\frac{1}{80};\ldots;\frac{1}{79}>\frac{1}{80};\frac{1}{80}=\frac{1}{80}\)

Do đó: \(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+\cdots+\frac{1}{80}=\frac{40}{80}=\frac12\) (2)

Từ (1),(2) suy ra \(B>\frac12+\frac12\)

=>B>1

mà \(1>\frac{39}{40}=A\)

nên B>A

Ta có:

A = 2 + 2² + 2³ + … + 2²⁰¹⁷

2A = 2.( 2 + 2² + 2³ + … + 2²⁰¹⁷)

2A = 2² + 2³ + 2⁴ + … + 2²⁰¹⁸

2A – A = (2² + 2³ + 2⁴ + … + 2²⁰¹⁸) – (2 + 2² + 2³ + … + 2²⁰¹⁷)

 Vậy  A = 2²⁰¹⁸ – 2

Ta có: A = 2 + 22 + 23 + … + 22017

2A = 2.( 2 + 22 + 23 + … + 22017)

2A = 22 + 23 + 24 + … + 22018

2A – A = (22 + 23 + 24 + … + 22018) – (2 + 22 + 23 + … + 22017)

A = 22018 – 2

Vậy A = 22018 – 2

tick cho mink nhé

😊

A = 2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰¹⁰

⇒ 2A = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰¹¹

⇒ A = 2A - A = (2 + 2² + 2³ + 2⁴ + ... + 2²⁰¹¹) - (2⁰ + 2¹ + 2² + 2³ + ... + 2²⁰¹⁰)

= 2²⁰¹¹ - 2⁰

= 2²⁰¹¹ - 1

= B

Vậy A = B

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