Question

Tính

​A=1/2+1/2^2+1/2^3+....+1/2^9

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A=

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Answer 1

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\)

\(2A=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\right)\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^8}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^8}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\right)\)

\(\Rightarrow A=1-\dfrac{1}{2^9}=\dfrac{2^9-1}{2^9}=\dfrac{511}{512}\)