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\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.......+\frac{1}{2^{2015}}\)
=>\(2A-A=1+\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{2015}}-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...........+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\right)\)
=>\(A=1-\frac{1}{2^{2016}}\)
=>\(A=\frac{2^{2016}-1}{2^{2016}}\)
Ta có : \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2016}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{2016}}\)
\(\Rightarrow2A-A=1-\frac{1}{2^{2016}}\)
\(\Rightarrow A=1-\frac{1}{2^{2016}}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)16
2A=\(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}+\frac{1}{2017}\)
2A-A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)-\(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}+\frac{1}{2^{2017}}\)
A=\(\frac{1}{2017}-\frac{1}{2}\)
Ta có :
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+..............+\dfrac{1}{2^{2015}}+\dfrac{1}{2^{2016}}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...........+\dfrac{1}{2^{2015}}\)
\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+..........+\dfrac{1}{2^{2015}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.......+\dfrac{1}{2^{2016}}\right)\)
\(\Rightarrow A=1-\dfrac{1}{2^{2016}}\)
\(\Rightarrow A=\dfrac{2^{2016}-1}{2^{2016}}\)
~ Học tốt ~
sai đề là cái chắc
ko chép đúng đề mà