Ta có: \(A=1+2+2^2+...+2^{2015}\)

\(2A=2\cdot\left(1+2+2^2+...+2^{2015}\right)\)

\(2A=2+2^2+2^3+...+2^{2016}\)

\(2A-A=2+2^2+...+2^{2016}-1-2-2^2-...-2^{2015}\)

\(A=2^{2016}-1\)

A không thể biết dưới dạng lũy thừa của 8 được 

A=22016−1

a) \(A=1+2+2^2+...+2^{80}\)

\(2A=2+2^2+2^3+...+2^{81}\)

\(2A-A=2+2^2+2^3+...+2^{81}-1-2-2^2-...-2^{80}\)

\(A=2^{81}-1\)

Nên A + 1 là:

\(A+1=2^{81}-1+1=2^{81}\)

b) \(B=1+3+3^2+...+3^{99}\)

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

\(3B-B=3+3^2+3^3+...+3^{100}-1-3-3^2-...-3^{99}\)

\(2B=3^{100}-1\)

Nên 2B + 1 là:

\(2B+1=3^{100}-1+1=3^{100}\)

2) 

a) \(2^x\cdot\left(1+2+2^2+...+2^{2015}\right)+1=2^{2016}\)

Gọi:

\(A=1+2+2^2+...+2^{2015}\)

\(2A=2+2^2+2^3+...+2^{2016}\)

\(A=2^{2016}-1\)

Ta có:

\(2^x\cdot\left(2^{2016}-1\right)+1=2^{2016}\)

\(\Rightarrow2^x\cdot\left(2^{2016}-1\right)=2^{2016}-1\)

\(\Rightarrow2^x=\dfrac{2^{2016}-1}{2^{2016}-1}=1\)

\(\Rightarrow2^x=2^0\)

\(\Rightarrow x=0\)

b) \(8^x-1=1+2+2^2+...+2^{2015}\)

Gọi: \(B=1+2+2^2+...+2^{2015}\)

\(2B=2+2^2+2^3+...+2^{2016}\)

\(B=2^{2016}-1\)

Ta có:

\(8^x-1=2^{2016}-1\)

\(\Rightarrow\left(2^3\right)^x-1=2^{2016}-1\)

\(\Rightarrow2^{3x}-1=2^{2016}-1\)

\(\Rightarrow2^{3x}=2^{2016}\)

\(\Rightarrow3x=2016\)

\(\Rightarrow x=\dfrac{2016}{3}\)

\(\Rightarrow x=672\)

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

2A= 2+2^2+2^3+....+2^101

A=2A-A=(2+2^2+2^3+...+2^101)-(1+2+2^2+...+2^100)

=>A=2^101-1

=>A+1=2^101

Vậy A+1 là một lũy thừa của 2.

A = 1 + 2 + 2^2 + ... + 2^100 

2A = 2 + 2^2 + 2^3 + ... + 2^101

2A - A = ( 2+ 2^2+ 2^3 + ... + 2^101 ) - ( 1+ 2 + 2^2 + ... + 2^100)

A = 2^101 - 1 

A + 1 = 2^101 

D

D

D

A=5⁰+5¹+...+5⁹⁹

=>5A=5+5²+5³+...+5¹⁰⁰

=>5A-A=4A=(5+5²+...+5¹⁰⁰)-(5⁰+5¹+...+5⁹⁹)=5¹⁰⁰-1

=>4A+1=5¹⁰⁰

\(A=8^2\cdot32^4\\ =2^6\cdot2^{20}\\ =2^{26}\\ B=27^3\cdot9^4\cdot243\\ =3^9\cdot3^{12}\cdot3^5\\ =3^{26}\)

`A=8^2*32^4=(2^3)^2*(2^5)^4=2^6*2^20=2^26`

`B=27^3*9^4*243=(3^3)^3*(3^2)^4*3^5=3^9*3^8*3^5=3^22`

\(A=8^2.32^4\)

\(=\left(2^3\right)^2.\left(2^5\right)^4\)

\(=2^6.2^{20}\)

\(=2^{26}\)

\(B=27^3.9^4.243\)

\(=\left(3^3\right)^3.\left(3^2\right)^4.3^5\)

\(=3^9.3^8.3^5\)

\(=3^{22}\)

a, \(A=1+2+2^2+2^3+...+2^{100}\)

=> \(2A=2+2^2+2^3+2^4+...+2^{101}\)

=> \(A=2A-A=2^{101}-1\)

=> \(A+1=2^{101}\)

b, \(B=3+3^2+3^3+...+3^{2005}\)

\(3A=3^2+3^3+3^4+....+3^{2006}\)

=> \(2A=3A-A=3^{2006}-3\)

=> \(2A+3=3^{2006}\)là lũy thừa của 3

=> Đpcm

a) Ta có: \(A=1+2+2^2+2^3+.....+2^{100}\)

\(\Rightarrow2A=2+2^2+2^3+........+2^{101}\)

Lấy 2A-A ta có: 

\(2A-A=\left(2+2^2+2^3+2^4+.....+2^{101}\right)\)\(-\left(1+2+2^2+2^3+.......+2^{100}\right)\)

\(\Rightarrow A=2^{101}-1\)

\(\Rightarrow A+1=2^{101}-1+1\)

\(\Rightarrow A+1=2^{101}\)

b) Ta có: \(B=3+3^2+3^3+.....+3^{2005}\)

\(\Rightarrow3B=3^2+3^3+3^4+.....+3^{2006}\)

\(\Rightarrow3B-B=\left(3^2+3^3+3^4+....+3^{2006}\right)\)\(-\left(3+3^2+3^3+......+3^{2005}\right)\)

\(\Rightarrow2B=3^{2006}-3\)

\(\Rightarrow2B+3=3^{2006}-3+3\)

\(\Rightarrow2B+3=3^{2006}\)

Vậy 2B+3 là lũy thừa của 3         ĐPCM

9+8^2+8^3+...8^50