Question

Tính:

\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2015}}+\dfrac{1}{2^{2016}}\)

Answer 1

Từ đề bài, ta có:

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}\)

\(\Rightarrow2A-A=A=1-\dfrac{1}{2^{2016}}\)

Vậy \(A=1-\dfrac{1}{2^{2016}}\)

Answer 2

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2015}}+\dfrac{1}{2^{2016}}\)

Ta có:

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2015}}+\dfrac{1}{2^{2016}}\)

\(\Rightarrow2A=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2015}}+\dfrac{1}{2^{2016}}\right)\)

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}\)

\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\right)\)

\(\Rightarrow A=1-\dfrac{1}{2^{2016}}\)

Vậy \(A=1-\dfrac{1}{2^{2016}}\)