A= (2+22)+(23+24)+...+(259+260)
A=2.(1+2)+23.(1+2)+...+259.(1+2)
A=2.3+23.3+...+259.3
A=3.(2+23+...+259)
Vì 3 chia hết cho 3 => 3.(2+23+...+259)  chia hết cho 3
=>A  chia hết cho 3
A= (2+22+23)+...+(258+259+260)
A=2.(1+2+22)+...+258.(1+2+22)
A=2.7+...+258.7
A=7.(2+...+258)
Vì 7  chia hết cho 7 =>7.(2+...+258)  chia hết cho 7

CHIA HẾT CHO 3 :

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

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

A=2.3+23.3+...+259.3

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

Vì 3 chia hết cho 3 => 3.(2+23+...+259) chia hết cho 3

=>A chia hết cho 3

dcv

Số số hạng của A:

90 - 1 + 1 = 90 (số)

Do 90 chia hết cho 3 nên có thể nhóm thành nhóm 3 số hạng

Ta có:

A = 2¹ + 2² + 2³ + ... + 2⁹⁰

= (2 + 2² + 2³) + (2⁴ + 2⁵ + 2⁶) + ... + (2⁸⁸ + 2⁸⁹ + 2⁹⁰)

= 2.(1 + 2 + 2²) + 2⁴.(1 + 2 + 2²) + ... + 2⁸⁸.(1 + 2 + 2²)

= 2.7 + 2⁴.7 + ... + 2⁸⁸.7

= 7.(2 + 2⁴ + ... + 2⁸⁸) ⋮ 7

Vậy A ⋮ 7

b) A = 2¹ + 2² + 2³ + ... + 2⁹⁰

⇒ 2A = 2² + 2³ + 2⁴ + ... + 2⁹¹

⇒ A = 2A - A = (2² + 2³ + 2⁴ + ... + 2⁹¹) - (2 + 2² + 2³ + ... + 2⁹⁰)

= 2⁹¹ - 2

Ta có: \(A=2+2^2+2^3+2^4+...+2^{99}+91\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{97}+2^{98}+2^{99}\right)+91\)

\(=2\cdot\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{97}\left(1+2+2^2\right)+91\)

\(=7\cdot\left(1+2^4+...+2^{97}\right)+7\cdot13\)

\(=7\cdot\left(1+2^4+...+2^{97}+13\right)⋮7\)(đpcm)

Ta có: \(A=2+2^2+2^3+2^4+...+2^{99}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{97}+2^{98}+2^{99}\right)\)

\(=2\cdot\left(1+2+2^2\right)+2^4\cdot\left(1+2+2^2\right)+...+2^{97}\left(1+2+2^2\right)\)

\(=\left(1+2+2^2\right)\cdot\left(2+2^4+...+2^{97}\right)\)

\(=7\cdot\left(2+2^4+...+2^{97}\right)⋮7\)(đpcm)

A = 2 + 2² + 2³ + 2⁴ + 2⁵..... + 2²³ + 2²⁴

=  (2 + 2² + 2³) + (2³ + 2⁴ + 2⁵) + ..... + (2²² + 2²³ + 2²⁴)

=  (2 + 2² + 2³) + 2² (2 + 2² + 2³) + .... + 2²². (2 + 2² + 2³)

= 14 + 2².14 + .... + 2²².14

= 14.(1 + 2² + ... + 2²²)

= 2.7.(1 + 2² + ... + 2²²) \(⋮\) 7

\(\Rightarrow A⋮7\)(ĐPCM)

1/

Tổng A là tổng các số hạng cách đều nhau 4 đơn vị.

Số số hạng: $(101-1):4+1=26$

$A=(101+1)\times 26:2=1326$

2/

$B=(1+2+2^2)+(2^3+2^4+2^5)+(2^6+2^7+2^8)+(2^9+2^{10}+2^{11})$

$=(1+2+2^2)+2^3(1+2+2^2)+2^6(1+2+2^2)+2^9(1+2+2^2)$

$=(1+2+2^2)(1+2^3+2^6+2^9)$

$=7(1+2^3+2^6+2^9)\vdots 7$

3/
$C=1+2+2^2+2^3+...+2^{99}$

$2C=2+2^2+2^3+2^4+...+2^{100}$

$\Rightarrow 2C-C=2^{100}-1$

$\Rightarrow C=2^{100}-1$

A = 2 + 22 + 23 + 24 + ... + 219 + 220

A = (2 + 22) + (23 + 24) +... + (219 + 220)

A = 2.(1+2) + 23.(1 + 2) +... + 219.(l + 2)

A = 2.3 + 23.3 +...+ 219.3 Do đó A chia hết cho 3

do đó A chia hết cho 3

\(A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)

\(=6+2^2.6+...+2^{98}.6=6\left(1+2^2+...+2^{98}\right)⋮6\)

\(A=2+2^2+2^3+...+2^{100}\)

\(=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)

\(=6+6.2^2+...+6.2^{98}\)

\(=6\left(1+2^2+...+2^{98}\right)⋮6\)

\(A=2+2^2+2^3+2^4+...+2^{100}\)

\(=2\cdot3+2^3\cdot3+...+2^{99}\cdot3\)

\(=6\left(1+2^2+...+2^{98}\right)⋮6\)