\(2B=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)
\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\right)\)
\(B=1-\frac{1}{2^{2016}}\)
phan kia tuong tu