Question

Tính

​A=1/2+1/2^2+...+1/2^9

Answer 1

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

\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.......+\dfrac{1}{2^8}\)

\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+.......+\dfrac{1}{2^8}\right)=-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^8}\right)\)

\(\Leftrightarrow A=1-\dfrac{1}{2^9}\)

Answer 2

Ta có: \(\dfrac{1}{2}A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

\(\Rightarrow A-\dfrac{1}{2}A=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)

\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^{10}}\)

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