Question

\(P=\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{2022}}\). So sánh P với \(\dfrac{1}{3}\)

Answer 1

\(P=\dfrac{1}{2^2}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2022}}\)

\(\Rightarrow4P=1+\dfrac{1}{2^2}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2020}}\)

\(\Rightarrow4P-P=1-\dfrac{1}{2^{2022}}\)

\(\Rightarrow3P=1-\dfrac{1}{2^{2022}}\)

\(\Rightarrow P=\dfrac{1}{3}-\dfrac{1}{3.2^{2022}}< \dfrac{1}{3}\)

Vậy \(P< \dfrac{1}{3}\)