a: \(=\left(1+2\right)+2^2\left(1+2\right)+...+2^{48}\left(1+2\right)\)

\(=3\left(1+2^2+...+2^{48}\right)⋮3\)

b: \(2^0+2^1+2^2+...+2^{101}\)

\(=\left(1+2+2^2\right)+...+2^{99}\left(1+2+2^2\right)\)

\(=7\left(1+...+2^{99}\right)⋮7\)

c: 2A=2+2^2+...+2^101

=>A=2^101-1

b) A=2+2²+2³+...+2²⁰

A=(2+2²)+(2³+2⁴)+...+(2¹⁹+2²⁰)

A=3.2+3.2³+...+3.2¹⁹

A=3.(2+2³+2⁵+...+2¹⁹)

⇒A⋮3

phần c) làm tương tự

a: \(A=1+2+2^2+...+2^{41}\)

=>\(2A=2+2^2+2^3+...+2^{42}\)

=>\(2A-A=2^{42}-1\)

=>\(A=2^{42}-1\)

b: \(A=\left(1+2\right)+2^2\left(1+2\right)+...+2^{40}\left(1+2\right)\)

\(=3\left(1+2^2+...+2^{40}\right)⋮3\)

\(A=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{39}\left(1+2+2^2\right)\)

\(=7\left(1+2^3+...+2^{39}\right)⋮7\)

a: Ta có: \(A=2+2^2+2^3+...+2^{20}\)

\(=2\left(1+2+2^2+...+2^{19}\right)⋮2\)

b: Ta có: \(A=2+2^2+2^3+...+2^{20}\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{19}\left(1+2\right)\)

\(=3\cdot\left(2+2^3+...+2^{19}\right)⋮3\)

c) tham khảo:

M = 2 + 22 + 23 + ... + 220
= ( 2 + 22 + 23 + 24 ) + ( 25 + 26 + 27 + 28 ) + ... + ( 217 + 218 + 219 + 220 )
= 2 . ( 1 + 2 + 22 + 23 ) + 25 . ( 1 + 2 + 22 + 23 ) + ... + 217 . ( 1 + 2 + 22 + 23 )
= 2 . 15 + 25 . 15 + ... + 217 .15
= 15 . 2 ( 1 + 24 + ... + 216 )
= 3 . 5 . 2 ( 1 + 24 + ... + 216 ) \(⋮\) 5

Lời giải:
a. 

$A=2(1+2^1+2^2+...+2^{19})\vdots 2$

b. 

$A=(2+2^2)+(2^3+2^4)+.....+(2^{19}+2^{20})$

$=2(1+2)+2^3(1+2)+....+2^{19}(1+2)$

$=2.3+2^3.3+...+2^{19}.3$

$=3(2+2^3+...+2^{19})\vdots 3$

c.

$A=(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+....+(2^{17}+2^{18}+2^{19}+2^{20})$

$=2(1+2+2^2+2^3)+2^5(1+2+2^2+2^3)+....+2^{17}(1+2+2^2+2^3)$

$=2.15+2^5.15+....2^{17}.15$
$=15(2+2^5+...+2^{17})$
$=5.3.(2+2^5+...+2^{17})\vdots 5$

a) \(A=2+2^2+2^3+...+2^{20}\)

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{19}+2^{20}\right)\)

\(A=2\cdot\left(1+3\right)+2^3\cdot\left(1+3\right)+...+2^{59}\cdot\left(1+3\right)\)

\(A=3\cdot\left(2+2^3+...+2^{59}\right)\)

Vậy A chia hết cho 3

________

\(A=2+2^2+2^3+...+2^{20}\)

\(A=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\)

\(A=2\cdot\left(1+4\right)+2^2\cdot\left(1+4\right)+...+2^{58}\cdot\left(1+4\right)\)

\(A=5\cdot\left(2+2^2+...+2^{58}\right)\)

Vậy A chia hết cho 5 

ta có :

A chia hết cho 15 nên A chia hết cho 3 và A chia hết cho 5

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

Ta có: 23a + 23b chia hết cho 23
=>\(7a+3b+16a+20b\) chia hết cho 23 
=>\(7a+3b+4\left(4a+5b\right)\)chia hết cho 23 

Theo đề bài: 7a + 3b chia hết cho 23

=> 4(4a + 5b) chia hết cho 23

Mà 4 không chia hết cho 23 nên 4a + 5b phải chia hết cho 23 (đpcm)