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) \(A=2+2^2+...+2^{120}\)

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

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

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

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

Mình chỉ làm được ý 3 thôi: 

A = 2¹ + 2² + 2³ + ................ + 2¹²⁰

Chứng minh chia hết cho 7

A = 2¹ + 2² + 2³ + ................ + 2¹²⁰

A = (2¹ + 2² + 2³) + (2⁴ + 2⁵ + 2⁶) + ................ + (2¹¹⁸ + 2¹¹⁹ + 2¹²⁰)

A = 2.(1 + 2 + 4) + 2⁴.(1 + 2 + 4) + ................. + 2¹¹⁸.(1 + 2 + 4)

A = 2.7 + 2⁴ . 7 + ................ + 2¹¹⁸.7

A = 7.(2 + 2⁴ + ........... + 2¹¹⁸)

Chứng minh chia hết cho 31

A = 2¹ + 2² + 2³ + ................ + 2¹²⁰ 

A = (2¹ + 2² + 2³ + 2⁴ + 2⁵) + (2⁶ + 2⁷ + 2⁸ + 2⁹ + 2¹⁰) + ................ + (2¹¹⁶ + 2¹¹⁷ + 2¹¹⁸ + 2¹¹⁹ + 2¹²⁰)

A = 2.(1 + 2 + 4 + 8 + 16) + 2⁶.(1 + 2 +4 + 8 + 16) + ............. + 2¹¹⁶.(1 + 2 + 4 + 8 + 16)

A = 2.31 + 2⁶.31 + ....... + 2¹¹⁶ . 31

A = 31.(2 + 2⁶ + ........... + 2¹¹⁶)

A = 2¹ + 2² + 2³ + ................ + 2¹²⁰

Chứng minh chia hết cho 7

A = 2¹ + 2² + 2³ + ................ + 2¹²⁰

A = (2¹ + 2² + 2³) + (2⁴ + 2⁵ + 2⁶) + ................ + (2¹¹⁸ + 2¹¹⁹ + 2¹²⁰)

A = 2.(1 + 2 + 4) + 2⁴.(1 + 2 + 4) + ................. + 2¹¹⁸.(1 + 2 + 4)

A = 2.7 + 2⁴ . 7 + ................ + 2¹¹⁸.7

A = 7.(2 + 2⁴ + ........... + 2¹¹⁸)

Chứng minh chia hết cho 31

A = 2¹ + 2² + 2³ + ................ + 2¹²⁰ 

A = (2¹ + 2² + 2³ + 2⁴ + 2⁵) + (2⁶ + 2⁷ + 2⁸ + 2⁹ + 2¹⁰) + ................ + (2¹¹⁶ + 2¹¹⁷ + 2¹¹⁸ + 2¹¹⁹ + 2¹²⁰)

A = 2.(1 + 2 + 4 + 8 + 16) + 2⁶.(1 + 2 +4 + 8 + 16) + ............. + 2¹¹⁶.(1 + 2 + 4 + 8 + 16)

A = 2.31 + 2⁶.31 + ....... + 2¹¹⁶ . 31

A = 31.(2 + 2⁶ + ........... + 2¹¹⁶)

A = 2 + 2² + ... + 2¹²⁰

Chứng minh chia hết cho 3

A = ( 2 + 2² ) + ( 2³ + 2⁴ ) + ... + ( 2¹¹⁹ + 2¹²⁰ )

= 2( 1 + 2 ) + 2³( 1 + 2 ) + ... + 2¹¹⁹( 1 + 2 )

= 2.3 + 2³.3 + ... + 2¹¹⁹.3

= 3( 2 + 2³ + ... + 2¹¹⁹ ) chia hết cho 3 ( đpcm )

Chứng minh chia hết cho 7

A = ( 2 + 2² + 2³ ) + ( 2⁴ + 2⁵ + 2⁶ ) + ... + ( 2¹¹⁸ + 2¹¹⁹ + 2¹²⁰ )

= 2( 1 + 2 + 2² ) + 2⁴( 1 + 2 + 2² ) + ... + 2¹¹⁸( 1 + 2 + 2² )

= 2.7 + 2⁴.7 + ... + 2¹¹⁸.7

= 7( 2 + 2⁴ + ... + 2¹¹⁸ ) chia hết cho 7 ( đpcm )

Chứng minh chia hết cho 15

A = ( 2 + 2² + 2³ + 2⁴ ) + ( 2⁵ + 2⁶ + 2⁷ + 2⁸ ) + ... + ( 2¹¹⁷ + 2¹¹⁸ + 2¹¹⁹ + 2¹²⁰ )

= 2( 1 + 2 + 2² + 2³ ) + 2⁵( 1 + 2 + 2² + 2³ ) + ... + 2¹¹⁷( 1 + 2 + 2² + 2³ )

= 2.15 + 2⁵.15 + ... + 2¹¹⁷.15

= 15( 2 + 2⁵ + ... + 2¹¹⁷ ) chia hết cho 15 ( đpcm )

1) Ta có: \(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{119}+2^{120}\right)\)

\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{119}\left(1+2\right)\)

\(A=3\left(2+2^3+...+2^{119}\right)\) chia hết cho 3

2) Ta có: \(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)

\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{118}\left(1+2+2^2\right)\)

\(A=7\left(2+2^4+...+2^{118}\right)\) chia hết cho 7

3) Ta có: \(A=\left(2+2^2+2^3+2^4\right)+...+\left(2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(A=2\left(1+2+2^2+2^3\right)+...+2^{117}\left(1+2+2^2+2^3\right)\)

\(A=15\left(2+2^5+...+2^{117}\right)\) chia hết cho 15

A=2+2^2+2^3+...+2^120

A=(2+2^2+2^3)+(2^4+2^5+2^6)...+(2^118+2^119+2^120)

A=2.(1+2+2^2)+2^4(1+2+2^2)+2^118(1+2+2^2)

A=2.7+2^4.7+...+2^118.7

Ta có A=2.7+2^4.7+...+2^118.7 chia hết cho 7

=>A=2+2^2+2^3+...+2^120 chia hết cho 7

A=2+2^2+...+2^120

=(2+2^2+2^3+2^4+2^5+2^6)+(2^7+2^8+2^9+2^10+2^11+2^12)+.....+(2^120+2^119+2^118+2^117+2^116+2^115)

=2(1+2+2^2+2^3+2^4+2^5)+2^7(1+2+2^2+2^3+2^4+2^5)+.....+2^115(1+2+2^2+2^3+2^4+2^5)

=2*63+2^7*63+...+2^115*63

=63(2+2^7+...+2^115) Vì 63 chia hết cho 7=>63(2+2^7+..+2^115) chia hết cho 7

=>A chia hết cho 7

chung minh rang A=7^0+7^1+7^2+7^3+................+7^30+7^31 chia het cho 25

ko biết

Minh lam cau A) thoi duoc hong

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

tất cả đều có trong câu hỏi tương tự

b)

A=(1+5+5²)+(5³+5⁴+5⁵)+...(5⁴⁰²+5⁴⁰³+5⁴⁰⁴)

A=31.1+31.5³+...+31.5⁴⁰²

A=31.(1+5³+...+5⁴⁰²)

=>A chia hết cho 31

=>Đâu phải con ma

\(A=7\left(1+7+7^2\right)+7^4\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)\)

\(=57\left(7+7^4+...+7^{118}\right)⋮57\)

\(A=7\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)\)

\(=57\left(7+...+7^{118}\right)⋮57\)

A =7(1+7+7₂)+7⁴(1+7+7₂)+...+7¹¹⁸(1+7+7²)A=7(1+7+7²)+74(1+7+7²)+...+7¹¹⁸(1+7+7²)

=57 (7+7⁴+...+71¹⁸)⋮57