Bài đâu ạ?

Cho B= 4+2^2 +2^3+...+2^2005 . Chứng tỏ rằng B là một lũy thừa của cơ số 2

 bài đây nhé bn

\(A=4+2^2+2^3+...+2^{2005}\)

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

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

\(A=4+2^2+2^3+...+2^{2006}-4-2^2-2^3-...-2^{2005}\)

\(A=2^{2006}\)

Vậy A là 1 luỹ thừa của cơ số 2

\(B=5+5^2+...+5^{2021}\)

\(5B=5^2+5^3+...+5^{2022}\)

\(5B-B=\left(5^2+5^3+...+5^{2022}\right)-\left(5+5^2+...+5^{2021}\right)\)

\(4B=5^{2022}-5\)

\(B=\frac{5^{2022}-5}{4}\)

\(B+8=\frac{5^{2022}-5}{4}+8\)

\(B+8=\frac{5^{2022}-5}{4}+\frac{32}{4}\)

\(B+8=\frac{5^{2022}-5+32}{4}\)

\(B+8=\frac{5^{2022}+27}{4}\)

=> B + 8 k thể là số b/ph của 1 số tn 

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

Chi tiết:

\(A=4+2^2+2^3+...+2^{2005}\\ 2A=4\cdot2+2^3+2^4+...+2^{2006}\\ 2A-A=\left(4\cdot2+2^3+2^4+...+2^{2006}\right)-\left(4+2^2+2^3+...+2^{2005}\right)\\ A=4\cdot2+2^{2006}-4-2^2=2^{2006}\left(Đpcm\right)\)

Không biết 

mình ko biết vì mình mới lớp 4 .....

\(A=2+2^2+2^3+...+2^{60}\)

\(\Leftrightarrow2A=2^2+2^3+2^4+...+2^{61}\)

\(\Leftrightarrow2A-A=2^{61}-2\)

\(\Leftrightarrow A+2=2^{61}-2+2\)

\(\Leftrightarrow A+2=2^{61}\left(đpcm\right)\)

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

\(\Rightarrow2A=8+2^3+...+2^{2022}\\ \Rightarrow2A-A=8+2^3+...+2^{2022}-4-2^2-...-2^{2021}\\ \Rightarrow A=8+2^{2022}-4-2^2=8-4-4+2^{2022}=2^{2022}\left(đpcm\right)\)

\(A=2^2+2^2+2^3+...+2^{2021}=2^3+2^4+...+2^{2021}=2^{2022}\left(đpcm\right)\)

A=2²+2²+2³+...+2²⁰²¹
A2=2(2²+2²+2³+...+2²⁰²¹)
A2=2³+2³+2⁴+...+2²⁰²²
A2-A= 2³+2³+2⁴+...+2²⁰²²-2²+2²+2³+...+2²⁰²¹
A=2³-2²+2²+2²⁰²²
A=8-8+2²⁰²²
A=2²⁰²²

 

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:

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

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

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

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

\(\Rightarrow3x=2016\)

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

\(\Rightarrow x=672\)