Question

\(\frac{1}{^{2^1}}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{^{2^{2003}}}+\frac{1}{2^{2004}}\)

Answer 1

Gọi \(A=\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{2003}}+\frac{1}{2^{2004}}\)

      \(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2002}}+\frac{1}{2^{2003}}\)

\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{2002}}+\frac{1}{2^{2003}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2004}}\right)\)

\(2A-A=1+\frac{1}{2}+...+\frac{1}{2^{2002}}+\frac{1}{2^{2003}}-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{2004}}\)

\(A=1+\frac{1}{2^{2004}}\)