Question

[maivananh]

tinh tong sau :

A= 1/2 +1/2^2+1/2^3+...+ 1/2^2015+1/2^2016

Answer 1

[Đặng Quốc Vinh]

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.......+\frac{1}{2^{2015}}\)

=>\(2A-A=1+\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{2015}}-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...........+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\right)\)

=>\(A=1-\frac{1}{2^{2016}}\)

=>\(A=\frac{2^{2016}-1}{2^{2016}}\)

Answer 2

[l҉o҉n҉g҉ d҉z҉]

Ta có : \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2016}}\)

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

\(\Rightarrow2A-A=1-\frac{1}{2^{2016}}\)

\(\Rightarrow A=1-\frac{1}{2^{2016}}\)