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

\(=3\left(1+2^2+...+2^{40}\right)⋮3\)

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

\(=7\left(1+2^3+...+2^{39}\right)⋮7\)

Bài 3: 

a) Ta có: \(C=2+2^2+2^3+...+2^{99}+2^{100}\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)

\(=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)

\(=31\cdot\left(2+2^6+...+2^{96}\right)⋮31\)(đpcm)

Bài 1: 

Ta có: \(A=3^{n+2}-2^{n+2}+3^n-2^n\)

\(=3^n\cdot9-2^n\cdot4+3^n-2^n\)

\(=3^n\left(9+1\right)-2^n\left(4+1\right)\)

\(=10\left(3^n-2^{n-1}\right)⋮10\)

Vậy: A có chữ số tận cùng là 0

Bài 2: 

Ta có: \(abcd=1000\cdot a+100\cdot b+10\cdot c+d\)

\(\Leftrightarrow abcd=1000\cdot a+96\cdot b+8c+2c+4b+d\)

\(\Leftrightarrow abcd=8\left(125a+12b+c\right)+\left(2c+4b+d\right)\)

mà \(8\left(125a+12b+c\right)⋮8\)

và \(2c+4b+d⋮8\)

nên \(abcd⋮8\)(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) \(A=1+2+2^2+2^3+...+2^{99}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{100}\)

\(\Rightarrow A=2A-A=2+2^2+...+2^{100}-1-2-2^2-...-2^{99}=2^{100}-1\)

b) \(A=1+2+2^2+...+2^{99}=\left(1+2+2^2+2^3\right)+2^4\left(1+2+2^2+2^3\right)+...+2^{96}\left(1+2+2^2+2^3\right)\)

\(=15+2^4.15+...+2^{96}.15=15\left(1+2^4+...+2^{96}\right)\)

\(=3.5\left(1+2^4+...2^{96}\right)\) chia hết cho 3 và 5

c) \(A=1+2+2^2+...+2^{99}\)

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

\(=1+2.7+...+2^{97}.7=1+7\left(2+...+2^{97}\right)\) chia 7 dư 1

=> A không chia hết cho 7

\(a,3A=3^2+3^3+...+3^{101}\\ \Rightarrow3A-A=3^2+3^3+...+3^{101}-3-3^2-...-3^{100}\\ \Rightarrow2A=3^{101}-3\\ \Rightarrow A=\dfrac{3^{101}-3}{2}\)

\(b,A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\\ A=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{99}\left(1+3\right)\\ A=\left(1+3\right)\left(3+3^3+...+3^{99}\right)\\ A=4\left(3+3^3+...+3^{99}\right)⋮4\)

\(A=3+\left(3^2+3^3+...+3^{100}\right)\\ A=3+3^2\left(1+3+...+3^{100}\right)\\ A=3+9\left(1+3+...+3^{100}\right).chia.9.dư.3\\ \Rightarrow A⋮̸9\)

a) rút gọn a

a = 3 + 3^3 + 3^2 + .. + 3^100

3a = 3^2 + 3^3 + .. + 3^101

3a - a = (3^2 + 3^3 + .. + 3^101) - (3 + 3^2 + .. + 3^100)

2a = 3^301 - 3

a = 3^101 - 3/2

b) chứng minh a chia hết cho 4 và k chia hết cho 9

a = 3 + 3^2 + .. + 3^100

a = (3 + 3^2) + .. + (3^99 + 3^100)

a = 3 (1 + 3) + .. + 3^99 (1 + 3)

a = 3.4 + .. + 3^99.4

a = (3 + .. + 3^99).4 ⋮ 4

vì 9 ⋮̸4

=> a ⋮̸9

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

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=2+2²+2³+...+2²⁰

a) Vì cơ số của mỗi lũy thừa là 2 => A chia hết cho 2

b)A=2+2²+2³+...+2²⁰ 

A=(2+2²)+(2³+2⁴)+...+(2¹⁹+2²⁰ )

A=(2+2²)+2³(2+2²)+...+2¹⁹(2+2² )

A=6+2³x6+...+2¹⁹x6

A=6x(1+2³+...+2¹⁹)

Vì 6 chia hết cho 3=> A chia hết cho 3

HT

Lời giải:

$A=(1+2)+(2^2+2^3)+....+(2^{2020}+2^{2021})$

$=3+2^2(1+2)+....+2^{2020}(1+2)$

$=3+3.2^2+....+3.2^{2020}$

$=3(1+2^2+....+2^{2020})\vdots 3$
Ta có đpcm.

A=\((1+2)+\left(2^2+2^3\right)+...+\left(2^{19}+2^{20}\right)\)

A=\(3.1+2^2\left(1+2\right)+...+2^{19}\left(1+2\right)\)

A=\(3.1+3.2^2+...+3.2^{19}\)

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

Vậy A\(⋮3\)

A=(1+2)+(22+23)+...+(219+220)(1+2)+(22+23)+...+(219+220)

A=3.1+22(1+2)+...+219(1+2)3.1+22(1+2)+...+219(1+2)

A=3.1+3.22+...+3.2193.1+3.22+...+3.219

A=3(1+22+...+219)3(1+22+...+219)⋮3⋮3

NÊN  A⋮3