Question

cho biểu thức: A=1+1/2 +1/2^2 +1/2^3 +.......+1/2^2017

Chứng tỏ A<1

Answer 1

Ta có : \(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\right)\)

            \(2A=2+\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+...+\frac{2}{2^{2017}}\)

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

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

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

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

Vậy A < 1 

Answer 2

\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)

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

\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2016}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\right)\)

\(A=2-\frac{1}{2^{2017}}\left(đpcm\right)\)

Answer 3

A<1.k mk nha.