Question

Giải giúp tớ với :1/2 +1/4+1/8+....+1/1024

Answer 1

\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)

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

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

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

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

Answer 2

Đặt:

\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)

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

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

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

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

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

\(A=1-\dfrac{1}{1024}=\dfrac{2023}{2024}\)

Answer 3

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)

\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)

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

\(\Rightarrow2A=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^{10}}\right)\)

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^9}\)

\(\Rightarrow2A-A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{10}}\)

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

\(\Rightarrow A=1-\dfrac{1}{1024}\)

\(\Rightarrow A=\dfrac{1023}{1024}\)

Vậy \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}=\dfrac{1023}{1024}\)