Question

A=1+2+\(2^2+2^3+.................+2^{2008}\)

B=\(2^{2009}\)

Answer 1

Ta có:

A = 1 + 2 + 2² + ... + 2²⁰⁰⁸

=> 2A = 2 + 2² + ... + 2²⁰⁰⁹

=> 2A - A = 2²⁰⁰⁹ - 1

=> A = 2²⁰⁰⁸ - 1 < 2²⁰⁰⁹ = B

Vậy B> A

Answer 2

 2A=2+2^2+...+2^2009

2A-A=(2+2^2+...+2^2009)-(1+2+...+2^2008)

A=2^2009-1

=>A<B

Answer 3

\(A=1+2+2^2+2^3+......+2^{2008}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{2009}\)

\(\Rightarrow2A-A=\left(2+2^2+2^3+.....+2^{2009}\right)-\left(1+2+2^2+2^3+....+2^{2008}\right)\)

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

Do \(2^{2009}-1< 2^{2009}\Rightarrow A< B\)

Answer 4

Sai hết;câu hỏi là tính A-B

Answer 5

Lộn B-A

Answer 6

\(A=1+2+2^2+2^3+...+2^{2007}+2^{2008}\)

\(\Leftrightarrow2A=2+2^2+2^3+...+2^{2008}+2^{2009}\)

\(\Leftrightarrow2A-A=\left(2+2^2+2^3+...+2^{2009}\right)-\left(1+2+2^2+2^3+...+2^{2008}\right)\)

\(\Leftrightarrow2A-A=2+2^2+2^3+...+2^{2009}-1-2-2^2-2^3-...-2^{2008}\)

\(\Leftrightarrow A=2^{2009}-1\)

\(B=2^{2009}\)

Vậy \(B-A=2^{2009}-\left(2^{2009}-1\right)\)

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

\(B-A=1\)