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a) \(A=\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{301}{3^{100}}\)
\(\Rightarrow3A=4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{100}}\)
\(\Rightarrow3A-A=\left(4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{99}}\right)-\left(\frac{4}{3}+\frac{7}{3^2}+...+\frac{301}{3^{100}}\right)\)
\(\Rightarrow2A=4+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{301}{3^{100}}\)
Đặt \(F=1+\frac{1}{3}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3F=3+1+...+\frac{1}{3^{97}}\)
\(\Rightarrow3F-F=\left(3+...+\frac{1}{3^{97}}\right)-\left(1+...+\frac{1}{3^{98}}\right)\)
\(\Rightarrow2F=3-\frac{1}{3^{98}}< 3\)
\(\Rightarrow F< \frac{3}{2}\)
\(\Rightarrow2A< 4+\frac{3}{2}\)
\(\Rightarrow2A< \frac{11}{2}\)
\(\Rightarrow A< \frac{11}{4}\left(đpcm\right)\)
2. \(B=\frac{11}{3}+\frac{17}{3^2}+\frac{23}{3^3}+...+\frac{605}{3^{100}}\)
\(\Rightarrow3B=11+\frac{17}{3}+\frac{23}{3^2}+...+\frac{605}{3^{99}}\)
\(\Rightarrow3B-B=\left(11+...+\frac{605}{3^{99}}\right)-\left(\frac{11}{3}+...+\frac{605}{3^{100}}\right)\)
\(\Rightarrow2B=11+2+\frac{2}{3}+...+\frac{2}{3^{98}}-\frac{605}{3^{100}}\)
Đặt \(D=2+\frac{2}{3}+...+\frac{2}{3^{98}}\)
\(\Rightarrow3D=6+2+...+\frac{2}{3^{97}}\)
\(\Rightarrow2D=6-\frac{2}{3^{98}}< 6\)( làm tắt )
\(\Rightarrow2D< 6\)
\(\Rightarrow D< 3\)
\(\Rightarrow2B< 11+3\)
\(\Rightarrow2B< 14\)
\(\Rightarrow B< 7\left(đpcm\right)\)
cho Q=\(\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{3n+1}{3^n}\)
n thuộc N*, chứng minh Q<11/4
\(M=\frac{1}{3^2}+\frac{2}{3^3}+...+\frac{10}{3^{11}}\)
\(\Rightarrow3M=\frac{1}{3}+\frac{2}{3^2}+...+\frac{10}{3^{10}}\)
\(\Rightarrow3M-M=\frac{1}{3}+\frac{2}{3^2}-\frac{1}{3^2}+\frac{3}{3^3}-\frac{2}{3^3}+...+\frac{10}{3^{10}}-\frac{9}{3^{10}}-\frac{10}{3^{11}}\)
\(\Rightarrow2M=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}-\frac{10}{3^{11}}=A-\frac{10}{3^{11}}\)
\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^9}+\frac{1}{3^{10}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^9}\)
\(\Rightarrow3A-A=1-\frac{1}{3^{10}}\)
\(\Rightarrow2A=1-\frac{1}{3^{10}}\Rightarrow A=\frac{1}{2}-\frac{1}{2.3^{10}}\Rightarrow A< \frac{1}{2}\)
\(\Rightarrow2M=A-\frac{10}{3^{11}}< A< \frac{1}{2}\)
\(\Rightarrow M< \frac{1}{4}\)
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(A< 1-\frac{1}{2010}=\frac{2009}{2010}< 1\)
\(\Rightarrow\)\(A< 1\) ( đpcm )
Vậy \(A< 1\)
Chúc bạn học tốt ~
Phân số \(\frac{n}{n+1}\) là phân số tối giản rồi bạn nhé
Hơi lâu nên đợi anh chút
\(D=\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{3n+1}{3^n}\)
\(\Rightarrow3D=4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{3n+1}{3^{n-1}}\)
\(\Rightarrow3D-D=\left(4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{3n+1}{3^{n-1}}\right)-\left(\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{3n+1}{3^n}\right)\)
\(\Rightarrow2D=4+1+\frac{1}{3}+...+\frac{1}{3^{n-2}}-\frac{3n+1}{3^n}\)
Đặt \(M=4+1+\frac{1}{3}+...+\frac{1}{3^{n-2}}\)
\(\Rightarrow3M=12+3+1+...+\frac{1}{3^{n-3}}\)
\(\Rightarrow3M-M=\left(12+3+1+...+\frac{1}{3^{n-3}}\right)-\left(4+1+\frac{1}{3}+...+\frac{1}{3^{n-2}}\right)\)
\(\Rightarrow2M=11-\frac{1}{3^{n-2}}< 11\)
\(\Rightarrow2M< 11\)
\(\Rightarrow M< \frac{11}{2}\)
\(\Rightarrow2D< \frac{11}{2}\)
\(\Rightarrow D< \frac{11}{4}\left(đpcm\right)\)