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Ta có \(I=\frac{11}{3}+\frac{17}{3^2}+...+\frac{605}{3^{100}}\left(1\right)\)
\(\Leftrightarrow3I=11+\frac{17}{3}+\frac{23}{3^2}+...+\frac{605}{3^{99}}\left(2\right)\)
Lấy \(\left(2\right)trừ\left(1\right)\)ta có
\(3I-I=11+\frac{6}{3}+\frac{6}{3^2}+...+\frac{6}{3^{99}}-\frac{605}{3^{100}}\)
\(\Leftrightarrow2I=11+6\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\frac{605}{3^{100}}\)
Xét \(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\left(3\right)\)
\(\Leftrightarrow3A=1+\frac{1}{3}+...+\frac{1}{3^{99}}\left(4\right)\)
Lấy\(\left(4\right)-\left(3\right)\)ta có
\(2A=1-\frac{1}{3^{100}}\)
\(\Leftrightarrow6A=3-\frac{1}{3^{99}}\)
Khi đó \(2I=11+3-\frac{1}{3^{99}}-\frac{605}{3^{100}}\)
\(\Leftrightarrow2I=14-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
Vì\(\frac{1}{3^{99}}+\frac{605}{3^{100}}>0\)
\(\Rightarrow2I< 14\)
\(\Leftrightarrow I< 7\left(đpcm\right)\)
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 )
\(\Rightarrow2D< 6\)
\(\Rightarrow D< 3\)
\(\Rightarrow2B< 11+3\)
\(\Rightarrow2B< 14\)
\(\Rightarrow B< 7\left(đpcm\right)\)
Ta có \(H=\frac{7}{3}+\frac{13}{3^2}+...+\frac{605}{3^{100}}\)
\(\Leftrightarrow3H=7+\frac{13}{3}+...+\frac{605}{3^{99}}\)
\(\Rightarrow2H=7+\frac{6}{3}+\frac{6}{3^2}+...+\frac{6}{3^{99}}-\frac{605}{3^{100}}\)
\(\Leftrightarrow2H=7+6\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\frac{605}{3^{100}}\)
Mà \(6\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)=3-\frac{1}{3^{99}}\)
\(\Rightarrow2H=7+3-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
\(\Leftrightarrow2H=10-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
Vì\(\frac{1}{3^{99}}+\frac{605}{3^{100}}>0\)
\(\Rightarrow2H< 10\)
\(\Leftrightarrow H< 5\left(1\right)\)
Ta có \(2H=10-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
Mà\(\frac{1}{3^{97}}+\frac{605}{3^{98}}< 22\)
hay\(\frac{1}{3^{99}}+\frac{605}{3^{98}}< \frac{22}{9}\)
\(\Rightarrow2H>10-\frac{22}{9}=\frac{68}{9}=2\cdot\left(3+\frac{7}{9}\right)\)
\(\Rightarrow H>3+\frac{7}{9}\left(2\right)\)
Từ \(\left(1\right)\left(2\right)\Rightarrowđpcm\)
