Question

B=1+2+2²+2³+...+2²⁰⁰⁸/1-2²⁰⁰⁹

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Answer 1

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Answer 2

\(B=\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)

Đặt: \(A=1+2+2^2+2^3+....+2^{2008}\)

\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{2009}\)

\(\Rightarrow2A-A=\left(2+2^2+2^3+2^4+...+2^{2009}\right)-\left(1+2+2^2+2^3+....+2^{2008}\right)\)

\(\Rightarrow2^{2009}-1\)

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

Answer 3

Đặt \(A=1+2+2^2+2^3+...+2^{2008}\)

\(\Rightarrow2A=2+2^2+2^3+2^4...+2^{2009}\)

\(\Rightarrow2A-A=2+2^2+2^3+2^4...+2^{2009}-1-2-2^2-2^3-...-2^{2008}\)

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

\(\Rightarrow B=\frac{2^{2009}-1}{1-2^{2009}}=\frac{-1\left(1-2^{2009}\right)}{1-2^{2009}}=-1\)

Answer 4

\(A=1+2+2^2+2^3+2^4+....+2^{2008}\)

\(2A=2+2^2+2^3+2^4+.....+2^{2009}\)

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

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