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

\(\Rightarrow2^2A=2^2.\left(\frac{1}{2^2}-\frac{1}{2^4}+...+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\right)\)

\(\Rightarrow4A=1-\frac{1}{2^2}+\frac{1}{2^4}-...-\frac{1}{2^{4n-2}}+\frac{1}{2^{4n}}-...-\frac{1}{2^{2002}}\)

\(\Rightarrow4A+A=\left(1-\frac{1}{2^2}+\frac{1}{2^4}-...-\frac{1}{2^{4n-2}}+\frac{1}{2^{4n}}-...-\frac{1}{2^{2002}}\right)+\left(\frac{1}{2^2}-\frac{1}{2^4}+...+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\right)\)

\(\Rightarrow5A=1-\frac{1}{2^{2004}}\)

Vì \(1-\frac{1}{2^{2004}}< 1.\)

\(\Rightarrow5A< 1\)

\(\Rightarrow A< \frac{1}{5}=0,2\)

\(\Rightarrow A< 0,2\left(đpcm\right).\)

Chúc bạn học tốt!

4S=\(\dfrac{4}{2^2}-\dfrac{4}{2^4}+\dfrac{4}{2^6}-...+\dfrac{4}{2^{4n-2}}-\dfrac{4}{2^{4n}}+...+\dfrac{4}{2^{2002}}-\dfrac{4}{2^{2004}}\)

4S=1-\(\dfrac{1}{2^2}+\dfrac{1}{2^4}-,...-\dfrac{1}{2^{2002}}\)

4S+S=1-\(\dfrac{1}{2^{2004}}\)

5S=\(\dfrac{2^{2004}-1}{2^{2004}}\)<1

\(\Rightarrow\)5S<1 hay S<\(\dfrac{1}{5}\)=0,2(đpcm)

Hoe..>>

Bài này mk gặp rồi nhờ cô giải hộ mà giờ mk quên mất tiêu rồi

Xin lỗi bn nha, mk k thể giúp đc rồi!

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