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a) Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)=> \(2.A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
=> \(2.A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\)=> \(1-A=1-\left(1-\frac{1}{2^{10}}\right)=\frac{1}{2^{10}}>\frac{1}{2^{11}}\)=> đpcm
b) Đặt B = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
Vì \(\frac{1}{2^2}
1.\(VT=\frac{c}{abc+ac+c}+\frac{b}{bc+b+abc}+\frac{abc}{abc+bc+b}=\frac{c}{ac+c+1}+\frac{1}{ac+c+1}+\frac{ac}{ac+c+1}=\frac{ac+c+1}{ac+c+1}=1=VP\)
Phần C đề thiếu
\(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(\Rightarrow3D-D=(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}})-\)\((\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}})\)
\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow6D-2D=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}\)
\(\Rightarrow4D=3-\frac{203}{3^{100}}\)
\(\Rightarrow D=\frac{3}{4}-\frac{\frac{203}{3^{100}}}{4}< \frac{3}{4}\left(đpcm\right)\)
\(\frac{1}{2^2}< \frac{1}{1\cdot2}\\ \frac{1}{3^2}< \frac{1}{2\cdot3}\\ \frac{1}{4^2}< \frac{1}{3\cdot4}\\ ...\\ \frac{1}{n^2}< \frac{1}{\left(n-1\right)\cdot n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{\left(n-1\right)\cdot n}\\ \Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\\ \Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}< 1\\ \Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\left(\text{với }n\in N;n\ge2\right)\)
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!
\(M=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{10}{3^{11}}\)
\(\Rightarrow3M=\frac{1}{3}+\frac{2}{3^2}+...+\frac{10}{3^{10}}\)
\(\Rightarrow3M-M=\left(\frac{1}{3}+\frac{2}{3^2}+...+\frac{10}{3^{10}}\right)-\left(\frac{1}{3^2}+\frac{2}{3^3}+...+\frac{10}{3^{11}}\right)\)
\(\Rightarrow2M=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}-\frac{10}{3^{11}}\)
Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}\)
\(\Rightarrow3A=1+\frac{1}{3}+...+\frac{1}{3^9}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^9}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}\right)\)
\(\Rightarrow2A=1-\frac{1}{3^{10}}< 1\)
\(\Rightarrow2A< 1\)
\(\Rightarrow A< \frac{1}{2}\)
\(\Rightarrow2M< \frac{1}{2}-\frac{10}{3^{11}}\)
\(\Rightarrow M< \frac{\frac{1}{2}-\frac{10}{3^{11}}}{2}\)
\(\Rightarrow M< \frac{1}{4}-\frac{1}{2.3^{11}}< \frac{1}{4}\)
\(\Rightarrow M< \frac{1}{4}\left(đpcm\right)\)
Đặt \(S=\frac{1}{10^2}+\frac{1}{11^2}+\frac{1}{12^2}+.....+\frac{1}{2014^2}\)
Ta có : \(S< \frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+.....+\frac{1}{2013.2014}\\\)
Đặt \(A=\frac{1}{9.10}+\frac{1}{10.11}+....+\frac{1}{2013.2014}\\ =>A=\left(\frac{1}{9}-\frac{1}{10}\right)+\left(\frac{1}{10}-\frac{1}{11}\right)+......+\left(\frac{1}{2013}-\frac{1}{2014}\right)\\ =>A=\frac{1}{9}-\frac{1}{2014}\\ \)
Vậy A<\(\frac{1}{9}\)
Mà A>S =>S<\(\frac{1}{9}\)
\(1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)
\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)(1)
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)
\(\Rightarrow2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{10}}\)
Thay A vào (1)
\(\Rightarrow1-\left(1-\frac{1}{2^{10}}\right)\)
\(=1-1+\frac{1}{2^{10}}=\frac{1}{2^{10}}\)
Ta có: 210 < 211
\(\Rightarrow\frac{1}{2^{10}}>\frac{1}{2^{11}}\)(đpcm)