Question

Cho \(P=\dfrac{1}{3}-\dfrac{1}{3^2}+\dfrac{1}{3^3}-\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}.\) CMR \(P< \dfrac{1}{4}\)

Answer 1

Ta có :

\(P=\dfrac{1}{3}-\dfrac{1}{3^2}+\dfrac{1}{3^3}-\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}\)

\(\Rightarrow3P=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+\dfrac{1}{3^4}-+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

\(\Rightarrow4P=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+\dfrac{1}{3^3}-+...-\dfrac{1}{3^{99}}+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}\)

\(\Rightarrow4P=1-\dfrac{1}{3^{100}}\)

\(\Rightarrow P=\dfrac{1}{4}.\left(1-\dfrac{1}{3^{100}}\right)\)

mà \(1-\dfrac{1}{3^{100}}< 1\)

\(\Rightarrow P< \dfrac{1}{4}\left(dpcm\right)\)