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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)
Đặt :
\(A=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-......+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+.........+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)
\(\Leftrightarrow2^2A=2^2\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-.......+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)
\(\Leftrightarrow4A=1-\dfrac{1}{2^2}+\dfrac{1}{2^4}-\dfrac{1}{2^6}+.......-\dfrac{1}{2^{4n-2}}+\dfrac{1}{2^{4n}}-.......-\dfrac{1}{2^{2002}}\)
\(\Leftrightarrow4A+A=\left(1-\dfrac{1}{2^2}+\dfrac{1}{2^4}-\dfrac{1}{2^6}+.......-\dfrac{1}{2^{4n-2}}+\dfrac{1}{2^{4n}}-......-\dfrac{1}{2^{2002}}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+......+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)
\(\Leftrightarrow5A=1-\dfrac{1}{2^{2004}}\)
\(\Leftrightarrow A=\left(1-\dfrac{1}{2^{2004}}\right):5\)
\(\Leftrightarrow A=\dfrac{1}{5}-\dfrac{1}{5}.\dfrac{1}{2^{2004}}< \dfrac{1}{5}=0,2\left(đpcm\right)\)
\(S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)
\(2^2S=2^2\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)
\(4S=1-\dfrac{1}{2^2}+...+\dfrac{1}{2^{4n}}+\dfrac{1}{2^{4n+2}}+...+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\)
\(4S+S=\left(1-\dfrac{1}{2^2}+...+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)
\(5S=1-\dfrac{1}{2^{2004}}< 1\Rightarrow S< \dfrac{1}{5}=0,2\)
a)
\(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...........-\dfrac{1}{2004^2}\)
\(\Leftrightarrow B=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+..............+\dfrac{1}{2004^2}\right)\)
Đặt :
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+.............+\dfrac{1}{2004^2}\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
..........................
\(\dfrac{1}{2004^2}< \dfrac{1}{2003.2004}\)
\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+..............+\dfrac{1}{2003.2004}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{2003}-\dfrac{1}{2004}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2004}\)
\(\Leftrightarrow A< \dfrac{2003}{2004}\)
\(\Leftrightarrow1-A< 1-\dfrac{2003}{2004}\)
\(\Leftrightarrow B< \dfrac{1}{2004}\left(đpcm\right)\)
b) \(S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-........+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+.......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\)
\(\Leftrightarrow2^2S=2^2\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+.....+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+....+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)
\(\Leftrightarrow4S=1-\dfrac{1}{2^2}+.......+\dfrac{1}{2^{4n}}-\dfrac{1}{2^{4n+2}}+.......+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\)
\(\Leftrightarrow4S+S=\left(1-\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+.......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)\)\(\Leftrightarrow5S=1-\dfrac{1}{2^{2004}}< 1\)
\(\Leftrightarrow S< \dfrac{1}{5}=0,2\)
\(\Leftrightarrow S< 0,2\left(đpcm\right)\)
cho mik hỏi mik ko hiểu tại sao từ 1/2^4n-2 khi nhân với 2^2 lại ra đc 1/2^4n vậy? Xin hãy giải đáp giùm mik
