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11 tháng 2

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

\(2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dots+\dfrac{1}{2^{2019}}\)

\(2A-A=\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dots+\dfrac{1}{2^{2019}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dots+\dfrac{1}{2^{2020}}\right)\)

\(A=2-\dfrac{1}{2^{2020}}\)

2 tháng 11 2023

a,     A = 1 + 3 + 32 + 33 + ... + 32000

    3.A =  3 + 32 + 33+ 33+... + 32001

    3A - A = 3 + 32 + 33 + ... + 32001 - (1 + 3 + 32 + 33 + ... + 32000)

     2A    = 3 + 32 + 33 + ... + 32001 -  1 - 3 - 32 - 33 - ... - 32000

     2A   = 32001 - 1 

       A   = \(\dfrac{3^{2001}-1}{2}\)

       

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 

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!

22 tháng 10 2023

a) P = 1 + 3 + 3² + ... + 3¹⁰¹

= (1 + 3 + 3²) + (3³ + 3⁴ + 3⁵) + ... + (3⁹⁹ + 3¹⁰⁰ + 3¹⁰¹)

= 13 + 3³.(1 + 3 + 3²) + ... + 3⁹⁹.(1 + 3 + 3²)

= 13 + 3³.13 + ... + 3⁹⁹.13

= 13.(1 + 3³ + ... + 3⁹⁹) ⋮ 13

Vậy P ⋮ 13

b) B = 1 + 2² + 2⁴ + ... + 2²⁰²⁰

= (1 + 2² + 2⁴) + (2⁶ + 2⁸ + 2¹⁰) + ... + (2²⁰¹⁶ + 2²⁰¹⁸ + 2²⁰²⁰)

= 21 + 2⁶.(1 + 2² + 2⁴) + ... + 2²⁰¹⁶.(1 + 2² + 2⁴)

= 21 + 2⁶.21 + ... + 2²⁰¹⁶.21

= 21.(1 + 2⁶ + ... + 2²⁰¹⁶) ⋮ 21

Vậy B ⋮ 21

c) A = 2 + 2² + 2³ + ... + 2²⁰

= (2 + 2² + 2³ + 2⁴) + (2⁵ + 2⁶ + 2⁷ + 2⁸) + ... + (2¹⁷ + 2¹⁸ + 2¹⁹ + 2²⁰)

= 30 + 2⁴.(2 + 2² + 2³ + 2⁴) + ... + 2¹⁶.(2 + 2² + 2³ + 2⁴)

= 30 + 2⁴.30 + ... + 2¹⁶.30

= 30.(1 + 2⁴ + ... + 2¹⁶)

= 5.6.(1 + 2⁴ + ... + 2¹⁶) ⋮ 5

Vậy A ⋮ 5

d) A = 1 + 4 + 4² + ... + 4⁹⁸

= (1 + 4 + 4²) + (4³ + 4⁴ + 4⁵) + ... + (4⁹⁷ + 4⁹⁸ + 4⁹⁹)

= 21 + 4³.(1 + 4 + 4²) + ... + 4⁹⁷.(1 + 4 + 4²)

= 21 + 4³.21 + ... + 4⁹⁷.21

= 21.(1 + 4³ + ... + 4⁹⁷) ⋮ 21

Vậy A ⋮ 21

e) A = 11⁹ + 11⁸ + 11⁷ + ... + 11 + 1

= (11⁹ + 11⁸ + 11⁷ + 11⁶ + 11⁵) + (11⁴ + 11³ + 11² + 11 + 1)

= 11⁵.(11⁴ + 11³ + 11² + 11 + 1) + 16105

= 11⁵.16105 + 16105

= 16105.(11⁵ + 1)

= 5.3221.(11⁵ + 1) ⋮ 5

Vậy A ⋮ 5

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 \(A=\dfrac{2^{2021}-1}{2^{2021}}\)

 

AH
Akai Haruma
Giáo viên
23 tháng 6 2023

1. 

$=153^2+2.47.153+47^2=(153+47)^2=200^2=40000$

2.

$=1,24^2-2.1,24.0,24+0,24^2=(1,24-0,24)^2=1^2=1$

3. Không phù hợp để tính nhanh 

4. 

$=15^8-(15^8-1)=1$

5.

$=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+...+(2019^2-2020^2)$

$=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+...+(2019-2020)(2019+2020)$

$=(-1)(1+2)+(-1)(3+4)+(-1)(5+6)+....+(-1)(2019+2020)$

$=(-1)(1+2+3+4+....+2019+2020)=(-1).2020(2020+1):2=-2041210$

DT
23 tháng 6 2023

6:

\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^4-1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^8-1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^{2020}-1\right)\left(2^{2020}+1\right)+1\\ =2^{4040}-1+1=2^{4040}\)

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

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

16 tháng 4 2022

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