K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

28 tháng 12 2022

loading...

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:

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

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

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

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

 

23 tháng 12 2023

A = 8⁸ + 2²⁰

= (2³)⁸ + 2²⁰

= 2²⁴ + 2²⁰

= 2²⁰.(2⁴ + 1)

= 2²⁰.17 ⋮ 17

Vậy A ⋮ 17

13 tháng 11 2021

A=2+22+23+...+299+2100A=2+22+23+...+299+2100

⇒2A=22+23+24+...+2100+2101⇒2A=22+23+24+...+2100+2101

⇒A=2101−2⇒A=2101−2

B=3+32+33+...+399+3100B=3+32+33+...+399+3100

⇒3B=32+33+34+...+3100+3101⇒3B=32+33+34+...+3100+3101

⇒2B=3101−3⇒2B=3101−3

⇒B=3101−32

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:

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

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)

20 tháng 4 2018

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

\(\Rightarrow A=\left(3+3^2+3^3+3^4\right)+.......+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

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

\(\Rightarrow A=3.40+.........+3^{97}.40\)

\(\Rightarrow A=40.\left(3+.......+3^{97}\right)\)

\(\Rightarrow A⋮40\)( 1 )

Vì \(A\)là tổng của các bậc lũy thừa của 3 nên \(A⋮3\)( 2 )

Từ ( 1 ) và ( 2 ) suy ra : \(A⋮40.3\)

\(\Rightarrow A⋮120\)

Vậy \(A⋮120\)( 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: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)