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:

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

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

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

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

\(\Rightarrow3x=2016\)

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

\(\Rightarrow x=672\)

`#3107`

\(A=1+2^1+2^2+2^3+...+2^{2015}\)

\(2A=2+2^2+2^3+2^4+...+2^{2016}\)

\(2A-A=\left(2+2^2+2^3+2^4+...+2^{2016}\right)-\left(1+2+2^2+2^3+...+2^{2015}\right)\)

\(A=2+2^2+2^3+2^4+...+2^{2016}-1-2-2^2-2^3-...-2^{2015}\)

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

Vậy, \(A=2^{2016}-1.\)

\(A=2^0+2^1+2^2+...+2^{2015}\)

\(2\cdot A=2^1+2^2+2^3+...+2^{2016}\)

\(A=2A-A=2^{2016}-2^0\)

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

\(A=2^1+2^2+2^3+...+2^{2016}\)

\(\Rightarrow A=2\left(1+2^1+2^2\right)+2^4\left(1+2^1+2^2\right)...+2^{2014}\left(1+2^1+2^2\right)\)

\(\Rightarrow A=2.7+2^4.7...+2^{2014}.7\)

\(\Rightarrow A=7\left(2+2^4...+2^{2014}\right)⋮7\)

\(\Rightarrow dpcm\)

A=(1+2+2^2)+2^3(1+2+2^2)+...+2^2013(1+2+2^2)+2^2016

=7(1+2^3+...+2^2013)+2^2016

Vì 2^2016 chia 7 dư 1

nên A chia 7 dư 1

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

\(B=2^{2018}-2^{2017}-2^{2016}-2^{2015}-2^{2014}\)

\(=>2B=2^{2019}-2^{2018}-2^{2017}-2^{2016}-2^{2015}\)

\(=>2B+B=2^{2019}-2^{2014}\)

\(=>B=\dfrac{2^{2019}-2^{2014}}{3}\)

\(A=2^{2017}-2^{2016}-2^{2015}-..........-2^5\)

\(\Leftrightarrow A=2^{2017}-\left(2^{2016}+2^{2015}+..........+2^5\right)\)

Đặt :

\(B=2^{2016}+2^{2017}+...........+2^5\)

\(\Leftrightarrow2B=2^{2017}+2^{2016}+..........+2^6\)

\(\Leftrightarrow2B-B=\left(2^{2017}+2^{2016}+.......+2^6\right)-\left(2^{2016}+2^{2015}+......+2^5\right)\)

\(\Leftrightarrow B=2^{2017}-2^5\)

\(\Leftrightarrow A=2^{2017}-\left(2^{2017}-2^5\right)\)

\(\Leftrightarrow A=2^{2017}-2^{2017}-2^5\)

\(\Leftrightarrow A=0+2^5\)

\(\Leftrightarrow A=32\)

A = 2²⁰¹⁷ - 2²⁰¹⁶ - 2²⁰¹⁵ - … - 2⁵

= 2²⁰¹⁷ - (2²⁰¹⁶ + 2²⁰¹⁵ + … + 2⁵)

Đặt E = 2²⁰¹⁶ + 2²⁰¹⁵ + … + 2⁵

2E = 2²⁰¹⁷ + 2²⁰¹⁶ + … + 2⁶

2E - E =(2²⁰¹⁷ - 2²⁰¹⁶ - … - 2⁶) - (2²⁰¹⁶ - 2²⁰¹⁵ - … - 2⁵)

E = 2²⁰¹⁷ - 2⁵

=> A = 2²⁰¹⁷ - (2²⁰¹⁷ - 2⁵)

= 2²⁰¹⁷ - 2²⁰¹⁷ + 2⁵

= 32

\(A=2^{2017}-2^{2016}-2^{2015}-...-2^5\)

\(\Rightarrow2A=2\left(2^{2017}-2^{2016}-2^{2015}-...-2^5\right)\)

\(2A=2^{2018}-2^{2017}-2^{2016}-...-2^6\)

\(\Rightarrow2A-A=A=\left(2^{2018}-2^{2017}-2^{2016}-...-2^6\right)-\left(2^{2017}-2^{2016}-2^{2015}-...-2^5\right)\)

\(A=2^{2018}-2^{2017}-2^{2017}-2^5=2^{2018}-2.2^{2017}-2^5\)

\(A=2^{2018}-2^{2018}-2^5=-2^5=-32\) vậy \(A=-32\)