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16 tháng 4 2022

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11 tháng 5 2023

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

\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2}.\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)\)\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\)

\(\Rightarrow A-\dfrac{1}{2}A=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\right)\)\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^{2022}}\)

\(\Rightarrow\dfrac{1}{2}A=\dfrac{2^{2021}-1}{2^{2022}}\)

\(\Rightarrow A=\dfrac{2^{2021}-1}{2^{2023}}.2=\dfrac{2^{2021}-1}{2^{2021}}\)

Vậy \(A=\dfrac{2^{2021}-1}{2^{2021}}\)

 

19 tháng 10 2023

\(A=2+2^2+2^3+...+2^{2020}+2^{2021}+2^{2022}\\=(2+2^2)+(2^3+2^4)+(2^5+2^6)+...+(2^{2021}+2^{2022})\\=2\cdot(1+2)+2^3\cdot(1+2)+2^5\cdot(1+2)+...+2^{2021}\cdot(1+2)\\=2\cdot3+2^3\cdot3+2^5\cdot3+...+2^{2021}\cdot3\\=3\cdot(2+2^3+2^5+..+2^{2021})\)

Vì \(3\cdot\left(2+2^3+2^5+...+2^{2021}\right)⋮3\)

nên \(A⋮3\).

\(Toru\)

19 tháng 10 2023

A=(2+22)+22(2+22)+...+22020(2+22)

A= 6.1+22.6+...+22020.6

A=6(1+22+...+22020) chia hết cho 3

vậy A chia hết cho 3

19 tháng 12 2021

\(P=\left(1+2\right)+2^2\left(1+2\right)+...+2^{2020}\left(1+2\right)\)

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

19 tháng 12 2021

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

7 tháng 5 2021

2A=2*(1+2+22+...+22020)=2+22+...+22021

2A-A=(1+2+22+...+22021)-(1+2+22+...+22020)

A=22021-1<2021

Giải:

A=1+2+22+23+...+22020

2A=2+22+23+24+...+22021

2A-A=(2+22+23+24+...+22021)-(1+2+22+23+...+22020)

A=22021-1

⇒A<22021

Chúc bạn học tốt!

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)

18 tháng 4 2022

A=1/2+1/22+1/23+...+1/22020+1/22021 > B=1/3+1/4+1/5+13/60

Ta có: �=12+122+123+124+...+122021+122022

⇒2�=1+12+122+123+...+122020+122021

⇒2�-�=(1+12+122+123+...+122020+122021)-(12+122+123+124+...+122021+122022)

⇒�=1-122022<1

⇒�<1   (1)

Lại có: �=13+14+15+1760

⇒�=1615

⇒�=1+115>1

⇒�>1    (2)

Từ (1) và (2)⇒�<�

Vậy 

27 tháng 8 2023

\(S=1+2+2^2+2^3+2^4+...+2^{2011}\)

\(\Rightarrow S=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{2009}\left(1+2+2^2\right)\)

\(\Rightarrow S=7+2^3.7+...+2^{2009}.7\)

\(\Rightarrow S=7\left(1+2^3+...+2^{2009}\right)⋮7\)

\(\Rightarrow dpcm\)

12 tháng 9 2021

\(B=2+2^2+2^3+2^4+...+2^{99}+2^{100}=2\left(1+2^2+2^3+2^4\right)+...+2^{96}\left(1+2^2+2^3+2^4\right)=2.31+2^6.31+...+2^{96}.31=31\left(2+2^6+...+2^{96}\right)⋮31\)

Cảm ơn bạn/chị nhé ạ!!!Thankyou very much!!!

 

1 tháng 9 2023

Bài 1

a, cm : A = 165 + 215 ⋮ 3

    A = 165 + 215

   A = (24)5 +  215

  A  = 220 + 215

 A  =  215.(25 + 1)

 A = 215. 33 ⋮ 3 (đpcm)

b,cm : B = 88 + 220 ⋮ 17

    B = (23)8 + 220 

    B =  216 + 220

    B = 216.(1 + 24)

    B = 216. 17 ⋮ 17 (đpcm)

 

 

  

1 tháng 9 2023

c, cm: C = 1 - 2 + 22 - 23 + 24 - 25 + 26 -...-22021 + 22022 : 6 dư 1

C=1+(-2+22-23+24- 25+26)+...+(-22017+22018-22019+22020-22021+22022)

C = 1 + 42 +...+ 22016.(-2 + 22 - 23 + 24 - 25 + 26)

C = 1 + 42+...+ 22016.42

C = 1 + 42.(20+...+22016)

42 ⋮ 6 ⇒ C = 1 + 42.(20+...+22016) : 6 dư 1 đpcm