Question

Cho A=\(\frac{1}{2^2}\)+\(\frac{1}{2^4}\)+\(\frac{1}{2^6}\)+\(\frac{1}{2^8}\)+.....+\(\frac{1}{2^{100}}\)

Chứng minh A<\(\frac{1}{3}\)

giúp mình vs ạ mk cần gấp mình cảm ơn trước ạ !!!!

Answer 1

Ta có  4A=\(1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\)

Trừ 4A cho A ta được 

3A = \(1-\frac{1}{2^{100}}\)=> 3A <1 => A<1/3 (đpcm)

Chúc bạn học tốt 

Answer 2

Ta có :\(A=\frac{1}{2^2}+...+\frac{1}{2^{100}}\)

\(2A=\frac{1}{2}+...+\frac{1}{2^{99}}\)

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

\(A=\frac{1}{2}-\frac{1}{2^{100}}\)

Lại có :

\(\frac{1}{3}=\frac{1}{2}-\frac{1}{6}\)

Vì \(\frac{1}{2^{100}}< \frac{1}{6}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{2^{100}}>\frac{1}{2}-\frac{1}{6}\)

\(\Rightarrow A>\frac{1}{3}\)

Vậy \(A>\frac{1}{3}\)(ĐPCM)