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`#3107.101107`

Gọi biểu thức trên là A

Ta có:

\(A=1+5^2+5^4+...+5^{40}\\ =1\cdot\left(1+5^2\right)+5^4\cdot\left(1+5^2\right)+...+5^{38}\cdot\left(1+5^2\right)\\ =\left(1+5^2\right)\cdot\left(1+5^4+...+5^{38}\right)\\ =26\cdot\left(1+5^4+...+5^{38}\right)\)

Vì \(26\cdot\left(1+5^4+...+5^{38}\right)\text{ }⋮\text{ }26\)

\(\Rightarrow A\text{ }⋮\text{ }26\)

_______

Gọi biểu thức trên là B

Ta có:

\(B=1+2^2+2^4+...+2^{100}\\ =1\cdot\left(1+2^2+2^4\right)+2^6\cdot\left(1+2^2+2^4\right)+...+2^{96}\cdot\left(1+2^2+2^4\right)\\ =\left(1+2^2+2^4\right)\cdot\left(1+2^6+...+2^{96}\right)\\ =21\cdot\left(1+2^6+...+2^{96}\right)\)

Vì \(21\cdot\left(1+2^6+...+2^{96}\right)\text{ }⋮\text{ }21\)

\(\Rightarrow B\text{ }⋮\text{ }21\)

_______

Gọi biểu thức trên là C

Ta có:

\(C=1+3^2+3^4+...+3^{100}\\ =1\cdot\left(1+3^2+3^4+3^6\right)+3^6\cdot\left(1+3^2+3^4+3^6\right)+...+3^{94}\cdot\left(1+3^2+3^4+3^6\right)\\ =\left(1+3^2+3^4+3^6\right)\cdot\left(1+3^6+...+3^{94}\right)\\ =820\cdot\left(1+3^6+...+3^{94}\right)\)

Vì \(820\cdot\left(1+3^6+...+3^{94}\right)\text{ }⋮\text{ }82\)

\(\Rightarrow C\text{ }⋮\text{ }82.\)

6 tháng 10 2023

a) \(A=1+5^2+5^4+5^6...+5^{40}\)

\(\Rightarrow A=\left(1+5^2\right)+5^4\left(1+5^2\right)+...+5^{38}\left(1+5^2\right)\)

\(\Rightarrow A=26+5^4.26+...+5^{38}.26\)

\(\Rightarrow A=26\left(1+5^4+...+5^{38}\right)⋮26\)

\(\Rightarrow1+5^2+5^4+5^6...+5^{40}⋮6\left(dpcm\right)\)

b) \(B=1+2^2+2^4+2^6+...+2^{100}\)

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

\(\Rightarrow B=21+2^6.21+...+2^{96}.21\)

\(\Rightarrow B=21\left(1+2^6+...+2^{96}\right)⋮21\)

\(\Rightarrow1+2^2+2^4+2^6+...+2^{100}⋮21\left(dpcm\right)\)

Bài C tương tự bạn tự làm nhé!

28 tháng 12 2022

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12 tháng 12 2021

Bài 1:

\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)

\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)

12 tháng 12 2021

Bài 2:

\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)

21 tháng 12 2019

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

 

6 tháng 11 2023

Chịu 🤭🤭🤭

17 tháng 10 2019

AH
Akai Haruma
Giáo viên
21 tháng 10 2023

Lời giải:
Đặt $A=1+2^2+2^4+....+2^{100}$

$A=(1+2^2+2^4)+(2^6+2^8+2^{10})+.....+(2^{96}+2^{98}+2^{100})$

$A=(1+2^2+2^4)+2^6(1+2^2+2^4)+....+2^{96}(1+2^2+2^4)$

$=(1+2^2+2^4)(1+2^6+....+2^{96})$

$=21(1+2^6+....+2^{96})\vdots 21$ 

Ta có đpcm.

23 tháng 12 2023

A = 8⁸ + 2²⁰

= (2³)⁸ + 2²⁰

= 2²⁴ + 2²⁰

= 2²⁰.(2⁴ + 1)

= 2²⁰.17 ⋮ 17

Vậy A ⋮ 17

AH
Akai Haruma
Giáo viên
31 tháng 12 2023

Câu 1: 

$A=(2+2^2)+(2^3+2^4)+(2^5+2^6)+....+(2^{2019}+2^{2020})$

$=2(1+2)+2^3(1+2)+2^5(1+2)+....+2^{2019}(1+2)$

$=(1+2)(2+2^3+2^5+...+2^{2019})=3(2+2^3+2^5+...+2^{2019})\vdots 3$

-----------------

$A=2+(2^2+2^3+2^4)+(2^5+2^6+2^7)+....+(2^{2018}+2^{2019}+2^{2020})$

$=2+2^2(1+2+2^2)+2^5(1+2+2^2)+....+2^{2018}(1+2+2^2)$

$=2+(1+2+2^2)(2^2+2^5+....+2^{2018})$

$=2+7(2^2+2^5+...+2^{2018})$

$\Rightarrow A$ chia $7$ dư $2$.

AH
Akai Haruma
Giáo viên
31 tháng 12 2023

Câu 2:

$B=(3+3^2)+(3^3+3^4)+....+(3^{2021}+3^{2022})$
$=3(1+3)+3^3(1+3)+...+3^{2021}(1+3)$

$=(1+3)(3+3^3+...+3^{2021})=4(3+3^3+....+3^{2021})\vdots 4$

-------------------

$B=(3+3^2+3^3)+(3^4+3^5+3^6)+...+(3^{2020}+3^{2021}+3^{2022})$

$=3(1+3+3^2)+3^4(1+3+3^2)+....+3^{2020}(1+3+3^2)$

$=(1+3+3^2)(3+3^4+...+3^{2020})=13(3+3^4+...+3^{2020})\vdots 13$ (đpcm)

28 tháng 12 2021

\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)

\(=4.\left(3+3^3+...+3^{2009}\right)\)

⇒ \(B\) ⋮ 4

29 tháng 12 2021

b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)

13 tháng 11 2023

1: \(A=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)

\(=2\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+...+2^{97}\right)\)

\(=30\left(1+2^4+...+2^{96}\right)⋮30\)

2:

\(B=3+3^2+3^3+...+3^{2022}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2021}+3^{2022}\right)\)

\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{2020}\left(3+3^2\right)\)

\(=12\left(1+3^2+...+3^{2020}\right)⋮12\)