\(A=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^3-...-\left(\dfrac{1}{2}\right)^{10}\\ \\2A=1-\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2-...-\left(\dfrac{1}{2}\right)^9\\ 2A-A=\left[1-\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2-...-\left(\dfrac{1}{2}\right)^9\right]-\left[\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^3-...-\left(\dfrac{1}{2}\right)^{10}\right]\\ A=1-\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^{10}\)
Ta có:
\(A=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^3-...-\left(\dfrac{1}{2}\right)^{10}\)
\(\Rightarrow\dfrac{1}{2}-A=\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{10}\)
\(\Rightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-A\right)=\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{11}\)
\(\Rightarrow\left(\dfrac{1}{2}-A\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-A\right)=\left[\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{10}\right]-\left[\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{11}\right]\)
\(\Rightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-A\right)=\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^{11}\)
\(\Rightarrow\dfrac{1}{2}-A=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^{10}\)
\(\Rightarrow A=\left(\dfrac{1}{2}\right)^{10}\)
Vì \(\left(\dfrac{1}{2}\right)^{10}\) > 0 nên A > 0
Vậy, A > 0
