Question

Tính :

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

Answer 1

Answer 2

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

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

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

2A=\(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2014}}+\dfrac{1}{2^{2015}}\)

2A-A=\(\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2014}}+\dfrac{1}{2^{2015}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2015}}+\dfrac{1}{2^{2016}}\right)\)

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

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

Answer 3

Ta có:

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

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

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

\(\Rightarrow\) 2A = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2014}}\)

\(\Rightarrow\) \(\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2014}}\right)\)- \(\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{2015}}\right)\)

= 2A - A

\(\Rightarrow\) A= 1 - \(\dfrac{1}{2^{2015}}\) = \(\dfrac{2^{2015}-1}{2^{2015}}\)

Vậy A = \(\dfrac{2^{2015}-1}{2^{2015}}\)