Question

Cho \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)

Chứng minh : \(A< 1\)

Answer 1

ta có : \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{99}\)

\(\Rightarrow\dfrac{1}{2}A=\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{100}\)

\(\Rightarrow\dfrac{1}{2}A=A-\dfrac{1}{2}A=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^{100}\) \(\Rightarrow A=2.\left(\dfrac{1}{2}A\right)=1-2\left(\dfrac{1}{2}\right)^{100}< 1\left(đpcm\right)\)

Answer 2

\(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)

\(\Rightarrow2A=2\cdot\left[\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\right]\)

\(2A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{97}+\left(\dfrac{1}{2}\right)^{98}\)

\(\Rightarrow A=2A-A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{98}-\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^3-...-\left(\dfrac{1}{2}\right)^{99}\)

\(A=1-\left(\dfrac{1}{2}\right)^{99}< 1\left(đpcm\right)\)