Question

Bài 1:Cho C=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\). Chứng minh C < \(\dfrac{1}{2}\)

Answer 1

\(C=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow3C=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)

\(\Rightarrow3C-C=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)

\(\Rightarrow2C=1-\dfrac{1}{3^{99}}\)

\(\Rightarrow C=\dfrac{1}{2}-\dfrac{1}{3^{88}.2}< \dfrac{1}{2}\)

Vậy \(C< \dfrac{1}{2}\)

Answer 2

\(C=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow3C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\)

\(\Rightarrow3C-C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...-\dfrac{1}{3^{99}}\)

\(\Rightarrow2C=1-\dfrac{1}{3^{99}}\)

\(\Rightarrow C=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

Mà \(1-\dfrac{1}{3^{99}}< 1\)

\(\Rightarrow C< \dfrac{1}{2}\) ( đpcm )

Answer 3

\(C=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)

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

\(3C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\)

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

\(C=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(C=\dfrac{1}{2}-\dfrac{1}{3^{99}.2}\)

\(C< \dfrac{1}{2}\)

\(\rightarrowđpcm\)

\(\)