\(3+3^2+3^3+...+3^{2012}\)

\(=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)

\(=3\left(1+3+3^2+3^3\right)+...+3^{2009}\left(1+3+3^2+3^3\right)\)

\(=40\left(3+...+3^{2009}\right)⋮40\)

​​

rrrrr

a) \(A=2+2^2+...+2^{120}\)

\(\Rightarrow A=\left(2+2^2\right)+...+\left(2^{119}+2^{120}\right)\)

\(\Rightarrow A=\left(2+2^2\right)+...+2^{118}.\left(2+2^2\right)\)

\(\Rightarrow A=6+...+2^{118}.6\)

\(\Rightarrow A=6.\left(1+...+2^{118}\right)⋮3\Rightarrow A⋮3\left(đpcm\right)\)

b) \(A=2+2^2+...+2^{120}\)

\(\Rightarrow A=\left(2+2^2+2^3\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)

\(\Rightarrow A=\left(2+2^2+2^3\right)+...+2^{117}.\left(2+2^2+2^3\right)\)

\(\Rightarrow A=14+...+2^{117}.14\)

\(\Rightarrow A=14.\left(1+...+2^{117}\right)⋮7\Rightarrow A⋮7\left(đpcm\right)\)

Ta có: A= 2 + 2² + 2³ + ... + 2⁶⁰= (2 +2²) + (2³+ 2⁴) + ... + (2⁵⁹ + 2⁶⁰).

             = 2 x (2 + 1) + 2³ x (2 + 1) + ... + 2⁵⁹ x (2 + 1).

             = 2 x 3 + 2³ x 3 + ... + 2⁵⁹ x 3.

             = 3 x ( 2 + 2³ + ... + 2⁵⁹).

Vì A = 3 x ( 2 + 2³ + ... + 2⁵⁹)  nên A chia hết cho 3.

           A= (2 +2² + 2³) + (2⁴ + 2⁵  + 2⁶) + ... + (2⁵⁸ + 2⁵⁹ + 2⁶⁰).

             = 2 x (1 + 2 + 2²) + 2⁴ x (1 + 2 + 2²) + ... + 2⁵⁸ x (1 + 2 + 2²).

             = 2 x 7 + 2⁴ x 7 + ... + 2⁵⁸ x 7.

             = 7 x ( 2 + 2⁴ + ... + 2⁵⁸).

Vì A = 7 x ( 2 + 2⁴ + ... + 2⁵⁸)  nên A chia hết cho 7.

  A= (2 +2² + 2³ + 2⁴) + (2⁵ + 2⁶  + 2⁷ + 2⁸) + ... + (2⁵⁷ + 2⁵⁸ + 2⁵⁹ + 2⁶⁰).

             = 2 x (1 + 2 + 2² + 2³) + 2⁵ x (1 + 2 + 2² + 2³) + ... + 2⁵⁷ x (1 + 2 + 2² + 2³).

             = 2 x 15 + 2⁵ x 15 + ... + 2⁵⁷ x 15.

             = 15 x ( 2 + 2⁴ + ... + 2⁵⁸).

Vì A = 15 x ( 2 + 2⁴ + ... + 2⁵⁸)  nên A chia hết cho 15.

1) Ta có : 11a + 22b + 33c

      = 11a + 11.2b + 11.3c

      = 11.(a + 2b + 3c) \(⋮\)11

=> 11a + 22b + 33c \(⋮\)11

2) 2 + 2² + 2³ + ... + 2¹⁰⁰

= (2 + 2²) + (2³ + 2⁴) + ... + (2⁹⁹ + 2¹⁰⁰)

= (2 + 2²) + 2².(2 + 2²) + ... + 2⁹⁸.(2 + 2²)

= 6 + 2².6 + ... + 2⁹⁸.6

= 6.(1 + 2² + .. + 2⁹⁸)

= 2.3.(1 + 2² + ... + 2⁹⁸) \(⋮\)3

=> 2 + 2² + 2³ + ... + 2¹⁰⁰ \(⋮\)3

3) Ta có:  abcabc = abc000 + abc

 = abc x 1000 + abc 

 = abc x (1000 + 1)

= abc x 1001 

= abc .7. 13.11 (1)

= abc . 7 . 13 . 11 \(⋮\)7 

=> abcabc \(⋮\)7

=> Từ (1) ta có : abcabc = abc x 7.11.13 \(⋮\)11

     => abcabc \(⋮\)11

=> Từ (1) ta có :  abcabc = abc . 7.11.13 \(⋮\)           13

    => => abcabc \(⋮\)13

1

.\(11a+22b+33c=11\left(a+2b+3c\right)⋮11\) 

\(\Rightarrow11a+22b+33c⋮11\left(đpcm\right)\) 

hc tốt

A=(2+2²+2³+2⁴+2⁴)+...+(2⁹⁷+2⁹⁸+2⁹⁹+2¹⁰⁰)

A=31+2⁵(2+2²+2³+2⁴+2⁴)+...+2⁹⁷(2+2²+2³+2⁴+2⁴)

A=31(1+2⁵+...+2⁹⁷)

=>2+2^2+2^3+.......+2^100) chia hết  cho 31

Ta có:

$A=1+2^2+2^4+2^6+...+2^{20}+2^{22}$

$=(1+2^2+2^4)+(2^6+2^8+2^{10})+(2^{12}+2^{14}+2^{16})+(2^{18}+2^{20}+2^{22})$

$=21+2^6\cdot(1+2^2+2^4)+2^{12}\cdot(1+2^2+2^4)+2^{18}\cdot(1+2^2+2^4)$

$=21+2^6\cdot21+2^{12}\cdot21+2^{18}\cdot21$

$=21\cdot(1+2^6+2^{12}+2^{18})$

Vì $21\vdots7$

nên $21\cdot(1+2^6+2^{12}+2^{18})\vdots7$

hay $A\vdots7$ (1)

Lại có:

$A=1+2^2+2^4+2^6+...+2^{20}+2^{22}$

$=(1+2^2+2^4+2^6)+(2^8+2^{10}+2^{12}+2^{14})+(2^{16}+2^{18}+2^{20}+2^{22})$

$=85+2^8\cdot(1+2^2+2^4+2^6)+2^{16}\cdot(1+2^2+2^4+2^6)$

$=85+2^8\cdot85+2^{16}\cdot85$

$=85\cdot(1+2^8+2^{16})$

Vì $85\vdots17$

nên $85\cdot(1+2^8+2^{16})\vdots17$

hay $A\vdots17$ (2)

Mặt khác: $(7,17)=1$ (3)

Từ (1); (2) và (3) $\Rightarrow A\vdots 7\cdot17=119$

$\text{#}Toru$

c)D=4+4²+4³+4⁴+...+4²⁰¹²

D=(4+4²)+(4³+4⁴)+...+(4²⁰¹¹+4²⁰¹²)

D=4.5+4³.5+4⁵.5+...+4²⁰¹¹.5

D=5.(4+4³+4²⁰¹¹)

=>D chia hết cho 5

=>ĐPCM

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