1. cho A = \(\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{301}{3^{100}}\)chứng minh: A< \(\frac{11}{4}\)2. cho B = \(\frac{11}{3}+\frac{17}{3^2}+\frac{23}{3^3}+...+\frac{605}{3^{100}}\)chứng minh: B<73. cho C = \(\frac{4}{3}+\frac{13}{3^2}+\frac{22}{3^3}+...+\frac{904}{3^{101}}\)chứng minh:...
Đọc tiếp
1. cho A = \(\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{301}{3^{100}}\)chứng minh: A< \(\frac{11}{4}\)
2. cho B = \(\frac{11}{3}+\frac{17}{3^2}+\frac{23}{3^3}+...+\frac{605}{3^{100}}\)chứng minh: B<7
3. cho C = \(\frac{4}{3}+\frac{13}{3^2}+\frac{22}{3^3}+...+\frac{904}{3^{101}}\)chứng minh: C<\(\frac{17}{4}\)
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 )
\(\Rightarrow2D< 6\)
\(\Rightarrow D< 3\)
\(\Rightarrow2B< 11+3\)
\(\Rightarrow2B< 14\)
\(\Rightarrow B< 7\left(đpcm\right)\)