Ta có:A=\(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2015}}\)
=>4A=\(1+\frac{1}{4}+...+\frac{1}{4^{2014}}\)
=>4A-A=(\(1+\frac{1}{4}+...+\frac{1}{4^{2014}}\))-(\(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2015}}\))
=>3A=1-\(\frac{1}{4^{2015}}\)
=>A=\(\frac{1}{3}-\frac{1}{3.4^{2015}}\frac{1}{4}\)
Vậy \(\frac{1}{4}