Question

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

Thực hiện phép tính trên

Answer 1

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

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

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

=>A=1-\(\dfrac{1}{^{ }2^{10}}=\dfrac{2^{10}-1}{2^{10}}=\dfrac{1023}{1024}\)

Answer 2

Đề là cái gì cơ? Tính à?

Answer 3

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

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

\(\Rightarrow2A-A=1-\dfrac{1}{2^{10}}\)

\(A=\dfrac{2^{10}-1}{2^{10}}=\dfrac{1023}{1024}\)

Vậy \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}=\dfrac{1023}{1024}\).