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

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

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

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

\(A=2^{2006}\)

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 

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

\(2A=8+2^3+2^4+...+2^{2006}\)

\(2A-A=\left(8+2^3+2^4+...+2^{2006}\right)-\left(4+2^2+2^3+...+2^{2005}\right)\)

\(A=8+2^{2006}-\left(4+2^2\right)\)

\(A=2^{2006}\)

suy ra đpcm. 

Cảm ơn bạn nha

a, Có 2B= 4.2+2^3+2^4+...+2^21

B=2B-B=(4.2+2^3+2^4+...+2^21)-(4+2^2+2^3+...+2^20) = 4.2 + 2^21 - 4 - 2^2 = 2^21

=>B là lũy thừa cơ số 2

Thank you!!!
nhưng mình ko hiểu gì cả???

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

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

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

Chi tiết:

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

a) B = 3 + 3² + ... + 3²⁰⁰⁵

3B = 3² + 3³ + ... + 3²⁰⁰⁶

3B - B = 3²⁰⁰⁶ - 3 

2B = 3²⁰⁰⁶ - 3

Theo bài ra : 2B + 3 = 3²⁰⁰⁶ - 3 + 3 = 3²⁰⁰⁶

a) A = 2²⁰⁰⁷-1 => A + 1  = 2²⁰⁰⁷

b) Do 2B = 3B - B = 3²⁰⁰⁶- 3 => 2B + 3 = 3²⁰⁰⁶

c) C = 4 + 2² + 2³+...+2²⁰⁰⁵ = 2² + 2³ + ...+ 2²⁰⁰⁵ + 4

2C - C = 2²⁰⁰⁶ - 2² + 4 =2²⁰⁰⁶ - 2² + 2² = 2²⁰⁰⁶