Question

Rut gon bieu thuc

A= 1+1/2+1/2^+1/2^+....+1/2^2017

Answer 1

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

\(\Rightarrow2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2016}}\)

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

\(-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}\right)\)

\(\Rightarrow A=2-\dfrac{1}{2^{2017}}=\dfrac{2^{2018}-1}{2^{2017}}\)

Answer 2

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

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

\(2A-A=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2016}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}\right)\)

\(A=2-2^{2017}\)