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)

\(A=1+2+2^2+2^3+....+2^{98}+2^{99}\\ \Leftrightarrow A=\left(1+2\right)+\left(2^2+2^3\right)+\left(2^4+2^5\right)+....+\left(2^{98}+2^{99}\right)\\ \Leftrightarrow A=3+2^2.\left(1+2\right)+2^4.\left(1+2\right)+....+2^{98}.\left(1+2\right)\\ \Leftrightarrow A=3+3.2^2+3.2^4+....+3.2^{98}\\ \Leftrightarrow A=3.\left(1+2^2+2^4+...+2^{98}\right)⋮3\)

A = 2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵ … + 2⁹⁹

^A=( 2⁰ + 2¹ + 2² + 2³ + 2⁴⁾ +( 2⁵ … + 2⁹⁹⁾

A= 2⁰.(2⁰ + 2¹ + 2² + 2³ + 2⁴)+2⁵.( 2⁵ … + 2⁹⁹⁾

A= 1. 31+ 2⁵.31… + 2⁹⁵.31

^A= 31. (1+2⁵...+2⁹⁵)

KL: ...... 

\(A=2^0+2^1+2^2+2^3+2^4+...+2^{99}=\left(2^0+2^1+2^2+2^3+2^4\right)+2^5\left(2^0+2^1+2^2+2^3+2^4\right)+...+2^{95}\left(2^0+2^1+2^2+2^3+2^4\right)=31+31.2^5+...+31.2^{95}=31\left(1+2^5+...+2^{95}\right)⋮31\)

A = 2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵ … + 2⁹⁹

A=( 2⁰ + 2¹ + 2² + 2³ + 2⁴⁾ +( 2⁵ … + 2⁹⁹⁾

A= 2⁰.(2⁰ + 2¹ + 2² + 2³ + 2⁴)+2⁵.( 2⁵ … + 2⁹⁹⁾

A= 1. 31+ 2⁵.31… + 2⁹⁵.31

A= 31. (1+2⁵...+2⁹⁵)

KL: ...... 

a, Ta có:

2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100

=  2 + 2 2 + 2 3 + 2 4 + 2 5 +...+ 2 96 + 2 97 + 2 98 + 2 99 + 2 100

= 2. 1 + 2 + 2 2 + 2 3 + 2 4 +...+ 2 96 1 + 2 + 2 2 + 2 3 + 2 4

=  2 . 31 + 2 6 . 31 + . . . + 2 96 . 31

=  2 + 2 6 + . . . + 2 96 . 31  chia hết cho 31

b, Ta có:

5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150

=  5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150

=  5 1 + 5 + 5 3 1 + 5 + 5 5 1 + 5 + . . . + 5 149 1 + 5

=  5 . 6 + 5 3 . 6 + 5 5 . 6 + . . . + 5 149 . 6

=  ( 5 + 5 3 + 5 5 + . . . + 5 149 ) . 6  chia hết cho 6

Ta lại có:

5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150

=  5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 +...+ 5 145 + 5 146 + 5 147 + 5 148 + 5 149 + 5 150  (có đúng 25 nhóm)

=  [ ( 5 + 5 4 ) + ( 5 2 + 5 5 ) + ( 5 3 + 5 6 ) ] + ... +  [ 5 145 + 5 148 ) + ( 5 146 + 5 149 ) + ( 5 147 + 5 150 ]

=  [ 5 ( 1 + 5 3 ) + 5 2 ( 1 + 5 3 ) + 5 3 ( 1 + 5 3 ) ] + ... +  [ 5 145 1 + 5 3 ) + 5 146 ( 1 + 5 3 ) + 5 147 ( 1 + 5 3 ]

=  ( 5 . 126 + 5 2 . 126 + 5 3 . 126 ) + ... +  ( 5 145 . 126 + 5 146 . 126 + 5 147 . 126 )

=  ( 5 + 5 2 + 5 3 ) . 126 +  ( 5 7 + 5 8 + 5 9 ) . 126 +  ... + ( 5 145 + 5 146 + 5 147 ) . 126

= 126.[ ( 5 + 5 2 + 5 3 ) + ( 5 7 + 5 8 + 5 9 ) + ... +  ( 5 145 + 5 146 + 5 147 ) ] chia hết cho 126.

Vậy  5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150  vừa chia hết cho 6, vừa chia hết cho 126

Chịu 🤭🤭🤭

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

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

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

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

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)

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