\(C=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2017}{4^{2017}}\)
\(4C=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2017}{4^{2016}}\)
\(4C-C=\left(1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2017}{4^{2016}}\right)-\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2017}{4^{2017}}\right)\)
\(3C=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2016}}-\frac{2017}{4^{2017}}\)
\(12C=4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2015}}-\frac{2017}{4^{2016}}\)
\(12C-3C=\left(4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2015}}-\frac{2017}{4^{2016}}\right)-\left(1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2016}}-\frac{2017}{4^{2017}}\right)\)
\(9C=4-\frac{2017}{4^{2016}}-\frac{1}{4^{2016}}+\frac{2017}{4^{2017}}\)
\(9C=4-\frac{8068}{4^{2017}}-\frac{4}{4^{2017}}+\frac{2017}{4^{2017}}\)
\(9C=4-\frac{10081}{4^{2017}}\)
=> 9C < 4
=> C < \(\frac{4}{9}\)< \(\frac{1}{2}\)(đpcm)
