S = 1 + 2 + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷

= (1 + 2) + (2² + 2³) + (2⁴ + 2⁵) + (2⁶ + 2⁷)

= (1 + 2) + 2²(1 + 2) + 2⁴(1 + 2) + 2⁶(1 + 2)

= (1 + 2)(1 + 2² + 2⁴ + 2⁶) 

= 3(1 + 2² + 2⁴ + 2⁶) \(⋮3\)(ĐPCM)

2S = 1 + 2 + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷ 

S = (1+2 ) + (2² + 2³ ) + (2⁴ + 2⁵ ) + (2⁶ +2⁷)

S = 3 + 2²(1+2) + 2⁴(1+2) + 2⁶(1+2)

S = 3+2².3 + 2⁴.3 + 2⁶ .3 

S = 3(1+2² + 2⁴ + 2⁶ ) \(⋮\) 3

=> đpcm

Vì 1+2=3 \(⋮\)3 =>(1+2)+2²+2³+2⁴+2⁵+2⁶+2⁷=>S \(⋮\)3

Vậy S \(⋮\)9

Nhớ k cho mình nhé !

A = 1+2+2²+...+2¹⁰

=> 2A = 2+2²+2³+...+2¹¹

=> 2A - A = (2+2²+2³+...+2¹¹) - (1+2+2²+...+2¹⁰)

=> A = 2¹¹ - 1

B = 1+3+3²+...+3¹⁰

=> 3B = 3+3²+3³+...+3¹¹

=> 3B - B = (3+3²+3³+...+3¹¹) - (1+3+3²+...+3¹⁰)

=> 2B = 3¹¹ - 1

=> B = \(\frac{3^{11}-1}{2}\)

A = 1 + 2 ¹ + 2 ² + 2 ³ + ... + 2 ⁹ + 2 ¹⁰

2A = 2 + 2 ² + 2 ³ + 2 ⁴ + ... + 2 ¹⁰  + 2 ¹¹

2A - A = ( 2 + 2 ² + 2 ³ + 2 ⁴ + ... + 2 ¹⁰  + 2 ¹¹ ) 

           - ( 1 + 2 ¹ + 2 ² + 2 ³ + ... + 2 ⁹ + 2 ¹⁰  )

   A     = 2 ¹¹  - 1

   A     = 2047

B = 1 + 3 ¹ + 3 ² + 3 ³ + ... + 3 ⁹  + 3 ¹⁰

3B = 3 ¹ + 3 ² + 3 ³ + 3 ⁴ + ... + 3 ¹⁰  + 3 ¹¹

3B - B= ( 3 ¹ + 3 ² + 3 ³ + 3 ⁴ + ... + 3 ¹⁰  + 3 ¹¹ )

            - ( 1 + 3 ¹ + 3 ² + 3 ³ + ... + 3 ⁹  + 3 ¹⁰ )

 2B    = 3 ¹¹ - 1

B       = \(\frac{3^{11}-1}{2}\)

B = 88573

`#3107.101107`

\(A = 2 + 2^2 + 2^3 + ... + 2^{2020} + 2^{2021} + 2^{2022}\)

\(= (2 + 2^2) + (2^3 + 2^4) + ... + (2^{2021} + 2^{2022})\)

\(=2(1+2) + 2^3(1 + 2) + ... + 2^{2021}(1 + 2)\)

\(=(1 + 2)(2 + 2^3 + ... + 2^{2021})\)

\(= 3(2 + 2^3 + ... + 2^{2021})\)

Vì \(3(2 + 2^3 + ... + 2^{2021})\) \(\vdots\) \(3\)

`\Rightarrow A \vdots 3`

Vậy, `A \vdots 3.`

=49-9+200=240

=49-9+200=240

ko hiểu

\(A=2+2^2+2^3+...+2^{10}\)

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)

\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)

\(A=2.3+2^3.3+...+2^9.3\)

\(A=3\left(2+2^3+...+2^9\right)\)

Ta có : \(3⋮3\Rightarrow A=3\left(2+2^3+...+2^9\right)⋮3\)

không

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