Question

Cmr \(\dfrac{1}{2^2}+\dfrac{1}{2^4}+.......+\dfrac{1}{2^{100}}< \dfrac{1}{3}\)

GIÚP MIK NHÉ,MIK Cần GẤP NGAY BÂY GIỜ

Answer 1

Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{100}}\)

\(\Rightarrow4A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{98}}\)

\(\Rightarrow4A-A=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{98}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{100}}\right)\)

\(\Rightarrow3A=\dfrac{1}{2}-\dfrac{1}{2^{98}}\)

\(\Rightarrow A=\dfrac{1}{6}-\dfrac{1}{2^{98}}:3\)

Vì \(\dfrac{1}{6}\) < \(\dfrac{1}{3}\) nên \(\dfrac{1}{6}-\dfrac{1}{2^{98}}:3\) < \(\dfrac{1}{3}\). \(\Rightarrow\) A < \(\dfrac{1}{3}\)

\(\Rightarrow\) ĐPCM