Question

[Trần Viết Phương]

 Chứng minh rằng:

Answer 1

[Hacker♪]

\(2^2\)\(A=2^2\)\(\left(\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-....+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}+....+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\right)\)

\(4A\)\(=1-\frac{1}{2^2}\)\(+\frac{1}{2^4}\)\(-\frac{1}{2^6}\)\(+.....-\frac{1}{2^{4n-2}}\)\(+\frac{1}{2^{4n}}\)\(-....-\frac{1}{2002}\)

\(4A+A=\)\(..........\)

\(5A=1-\frac{1}{2^{2004}}\)

\(A=\left(1-\frac{1}{2^{2004}}\right)\)\(:5\)

\(A=\frac{1}{5}\)\(-\frac{1}{5}\)\(x\frac{1}{2^{2004}}\)\(< \frac{1}{5}\)\(\left(=0,2\right)\)

\(=>A< 0,2\)

Answer 2

[Member lỗi thời :&gt;&gt;]

Ta có :

\(S=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-\frac{1}{2^8}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\)

\(\Rightarrow2^2S=1-\frac{1}{2^2}+\frac{1}{2^4}-\frac{1}{2^6}+...+\frac{1}{2^{2000}}-\frac{1}{2^{2002}}\)

\(\Rightarrow4S+S=\left(1-\frac{1}{2^2}+\frac{1}{2^4}-\frac{1}{2^6}+...+\frac{1}{2^{2000}}-\frac{1}{2^{2002}}\right)\)

\(+\left(\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-\frac{1}{2^8}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\right)\)

\(\Rightarrow5S=1-\frac{1}{2^{2004}}< 1\)\(\Rightarrow S=\frac{1-\frac{1}{2^{2004}}}{5}< \frac{1}{5}=0,2\)