Question

CMR: \(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}< \dfrac{1}{2}\)

Answer 1

Answer 2

Đặt A \(=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}\)

\(\Rightarrow3A=1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\)

\(\Rightarrow3A-A=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}\right)\)

\(\Rightarrow2A=1+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)-\dfrac{100}{3^{100}}\)

\(\Rightarrow2A=1+\dfrac{1}{2}-\dfrac{1}{3^{99}\times2}-\dfrac{100}{3^{100}}\)

\(\Rightarrow A=\dfrac{3}{4}-\dfrac{1}{3^{99}\times4}-\dfrac{50}{3^{100}}< \dfrac{3}{4}\)

Vậy ...

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