`A=1+2^2 +2^3 +...+2^10`

`2A=2+2^3 +2^4 +...+2^11`

`A=2+2^3 +2^4 +...+2^11 -1-2^2 -2^3 -...-2^10`

`A=2+2^11 -1-2^2`

`A=2+2048-1-4`

`A=2045`

Đặt: \(A=1+2^2+2^3+...+2^{10}\)

\(\Rightarrow2A=2\cdot\left(1+2^2+2^3+...+2^{10}\right)\)

\(\Rightarrow2A=2+2^3+2^4+...+2^{11}\)

\(\Rightarrow2A-A=\left(2+2^3+2^4+...+2^{11}\right)-\left(1+2^2+2^3+...+2^{10}\right)\)

\(\Rightarrow A=2+2^3+2^4+...+2^{11}-1-2^2-2^3-...-2^{10}\)

\(\Rightarrow A=\left(2^3-2^3\right)+\left(2^4-2^4\right)+...+\left(2^{10}-2^{10}\right)+\left(2+2^{11}-1-2^2\right)\)

\(\Rightarrow A=0+0+0+...+2+2^{11}-1-2^2\)

\(\Rightarrow A=2+2^{11}-1-4\)

\(\Rightarrow A=2^{11}-3\)

A = 1 + 2 + 2² + 2³ + 2⁴ + ..... + 2²⁰²¹

2A = 2 + 2² + 2³ + 2⁴ + 2⁵ + ..... + 2²⁰²²

2A - A = ( 2 + 2² + 2³ + 2⁴ + 2⁵ + ..... + 2²⁰²² ) - ( 1 + 2 + 2² + 2³ + 2⁴ + ..... + 2²⁰²¹ )

A = 2²⁰²² - 1

cảm ơn bạn nhé

mà bạn ơi kết quả cuối cùng là A=2022-1 

mà bạn ơi mik nhầm mik bảo nè thế kết quả cuối cùng là A=2^2021 -1 à bạn

a: ${}^{}\cdot {}^{}={}^{}={}^{}$

b: 64⋅83=24⋅34⋅29=213⋅3464⋅83=24⋅34⋅29=213⋅34

c: $16\cdot 81={}^{}$

d: ${}^{}\cdot {}^{}={}^{}$

\(B=\dfrac{1+2+2^2+.............................+2^{2008}}{1-2^{2009}}\)

Đặt \(N=1+2+2^2+..........+2^{2008}\)

\(\Rightarrow2N=2+2^2+2^3+.................+2^{2009}\)

2N-N=\(\left(2+2^2+2^3+............+2^{2009}\right)-\left(1+2+2^2+............+2^{2008}\right)\)

\(N=2^{2009}-1\)

Thay N vào B được

\(B=\dfrac{1-2^{2009}}{2^{2009}-1}=-1\)

Vậy .........................

Chúc bn học tốt

Giải:

\(B=\dfrac{1+2+2^2+2^3+...+2^{2018}}{1-2^{2009}}\) 

Đặt \(A=1+2+2^2+2^3+...+2^{2008}\) 

\(2A=2+2^2+2^3+2^4+...+2^{2009}\) 

\(2A-A=\left(2+2^2+2^3+2^4+...+2^{2009}\right)-\left(1+2+2^2+2^3+...+2^{2008}\right)\) 

\(A=2^{2009}-1\) 

\(\Rightarrow B=\dfrac{2^{2009}-1}{1-2^{2009}}=-1\)

S=1+2+2²+2³+...+2²⁰

2S=2+2²+2³+2⁴+...+2²¹  

=>S=2S-S=2²¹-1C

Vậy S=2²¹-1

\(S=1+2+2^2+2^3+...+2^{20}\)

\(\Rightarrow2S=2+2^2+2^3+...+2^{21}\)

\(\Rightarrow2S-S=\left(2+2^2+...+2^{21}\right)-\left(1+2+...+2^{20}\right)\)

\(\Rightarrow S=2^{21}-1\)

S=1+2+2^2+....+2^20

=>2S=2+2^2+2^3+...+2^21

=>2S-S=2^21-1

=>S=2^21-1

M = 2²⁰¹⁰-(2²⁰⁰⁹ + 2²⁰⁰⁸+....+2¹+2⁰

Đặt A =( 2²⁰⁰⁹+2²⁰⁰⁸+...2¹ +2⁰)

Suy ra 2A = 2²⁰¹⁰+2²⁰⁰⁹+2²⁰⁰⁸+...2²+2

Suy ra 2A-A = ( 2²⁰¹⁰+2²⁰⁰⁹+2²⁰⁰⁸+...+2²+2) - (2²⁰⁰⁹+ 2²⁰⁰⁸+...+2¹+2⁰)

Suy ra A= 2²⁰¹⁰-2⁰

Suy ra M = 2²⁰¹⁰-A=2^{2010 -} 2²⁰¹⁰+2⁰=1

Vậy M=1

Đúng nha

tui biet cach lam rui

hay qua

ta có 1/2mũ 2 +1/3 mũ 2+1/4 mũ 2+...+1/100 mũ 2=1/2.2+1/3.3+1/4.4+...+1/100.100<1/2.3+1/3.4+1/4.5+...+1/99.100+1/100.101=1/2.3-1/100.101=1/6-1/10100=tự tính nhé

Chỗ ... là gì bạn

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{1000^2}\)

\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{999.1000}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)

\(=1-\dfrac{1}{1000}=\dfrac{999}{1000}< 1\left(đpcm\right)\)

\(A=2^0+2^1+2^2+2^3+2^4+2^5+\dots+2^{100}\\=(2^1+2^2)+(2^3+2^4)+(2^5+2^6)+\dots+(2^{99}+2^{100})+2^0\\=2\cdot(1+2)+2^3\cdot(1+2)+2^5\cdot(1+2)+\dots+2^{99}\cdot(1+2)+1\\=2\cdot3+2^3\cdot3+2^5\cdot3+\dots+2^{99}\cdot3+1\\=3\cdot(2+2^3+2^5+\dots+2^{99})+1\)

Vì \(3\cdot(2+2^3+2^5+\dots+2^{99})\vdots3\)

\(\Rightarrow 3\cdot(2+2^3+2^5+\dots+2^{99})+1\) chia \(3\) dư 1

hay số dư của phép chia \(A\) cho \(3\) là \(1\).

A=2^0 + 2^1 + 2^2 + 2^3 + 2^4 + ....+2^100

A=1 + 2^1 + 2^2 + 2^3 + 2^4 + ....+2^100

A=1 + (2^1 + 2^2) + (2^3 + 2^4) + ....+(2^99 + 2^100)

A=1 + 2.(1+2) + 2^3.(1+2)+....+2^99.(1+2)

A=1 + 2 . 3 + 2^3 . 3 +....+2^99 . 3

A=1 +3 .(2+2^3+..+2^99)

=> A:3 dư 1