Question

tính 

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2018}}\)

giúp mk đi làm ơn

Answer 1

ta có:

\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2017}}\)

\(\Rightarrow2A-A=2-\frac{1}{2^{2018}}\)

\(\Rightarrow A=\frac{2^{2019}-1}{2^{2018}}\)

Answer 2

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2018}}\)

\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{2017}}\)

\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+........+\frac{1}{2^{2017}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{2018}}\right)\)

\(\Rightarrow A=2-\frac{1}{2^{2018}}\)

\(\Rightarrow A=\frac{2^{2019}-1}{2^{2018}}\)

Answer 3

\(A=1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{2017}}+\frac{1}{2^{2018}}\)

\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{2017}}+\frac{1}{2^{2018}}\right)\)

\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{2016}}+\frac{1}{2^{2017}}\)

\(2A-A=2-\frac{1}{2^{2018}}\)

\(A=\frac{2^{2019}-1}{2^{2018}}\)