Đặt \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{100}}\)
\(\Rightarrow2A=2\times\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{100}}\right)\)
\(\Rightarrow2A=1+\dfrac{1}{2}+....+\dfrac{1}{2^{99}}\)
\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{100}}\right)\)\(\Rightarrow A=1-\dfrac{1}{2^{100}}\)
Vậy \(A=1-\dfrac{1}{2^{100}}\)
~ Học tốt ~
\(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\\ =\left(2-1\right)\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\right)\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\\ =1-\dfrac{1}{2^{100}}\)
