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![](https://rs.olm.vn/images/avt/0.png?1311)
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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![](https://rs.olm.vn/images/avt/0.png?1311)
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!
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
S = \(...\)
=> \(S.2^2=1-\frac{1}{2^2}+\frac{1}{2^4}-...+\frac{1}{2^{2000}}-\frac{1}{2^{2002}}\)
=> \(S.4+S=\left(1-\frac{1}{2^2}+\frac{1}{2^4}-...-\frac{1}{2^{2002}}\right)+\left(\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-...-\frac{1}{2^{2004}}\right)\)
=> \(5S=1-\frac{1}{2^{2004}}<1\)
=> \(S<1:5=0,2\left(đpcm\right)\)
Vậy S < 0,2.