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D=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^2}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\)
D=\(\frac{1}{3}+\frac{101}{3^{101}}\)
D=\(\frac{1}{3}\)
\(\frac{1}{3}và\frac{3}{4}\)
\(\frac{1}{3}=\frac{4}{12}\)
\(\frac{3}{4}=\frac{9}{12}\)
Vì\(\frac{4}{12}< \frac{9}{12}Vậy\frac{1}{3}< \frac{3}{4}\)
Bài 2:
M = 1/2.3/4.5/6...99/100
Ta có: \(\frac{a}{b}\) = 1 - \(\frac{b-a}{b}\) (a; b; n ∈ N* và b > a)
\(\frac{a+n}{b+n}\) = 1 - \(\frac{b-a}{b+n}\)
\(\frac{a}{b}\) < \(\frac{a+n}{b+n}\)
Áp dụng công thức trên ta có:
\(\frac12<\frac{1+1}{2+1}=\frac23\)
\(\frac34<\frac{3+1}{4+1}=\frac45\)
\(\frac56\) < \(\frac{5+1}{6+1}\) = \(\frac67\)
............................
\(\frac{99}{100}\) < \(\frac{99+1}{100+1}\) = \(\frac{100}{101}\)
Cộng vế với vế ta có:
M = \(\frac12\).\(\frac34\).\(\frac56\)...\(\frac{99}{100}\) < \(\frac23\).\(\frac45\)..\(\frac{100}{101}\) = N
M < N (đpcm)
b; M.N = \(\frac12\).\(\frac34\).\(\frac56\)...\(\frac{99}{100}\).\(\frac23\).\(\frac45\)..\(\frac{100}{101}\)
M.N = \(\frac{1.3.5\ldots99}{3.5\ldots101}\). \(\frac{2.4.6\ldots100}{2.4.6\ldots100}\)
M.N = 1/100.101
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 )
\(\Rightarrow2D< 6\)
\(\Rightarrow D< 3\)
\(\Rightarrow2B< 11+3\)
\(\Rightarrow2B< 14\)
\(\Rightarrow B< 7\left(đpcm\right)\)