Chứng minh:
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{101}{3^{101}}
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8 tháng 8 2020
Giúp mình nha. Bài cuối cùng của đề toán dài 36 bài của mình đó
8 tháng 8 2020
\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1\)
Nên từ đây => \(A< 1\) (ĐPCM)
DT
A =\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{101}{3^{101}}\)
Chứng minh A < \(\frac{3}{4}\)
0
18 tháng 3 2018
Đặt \(S=\frac{1}{3}+\frac{2}{3^2}+.......+\frac{101}{3^{101}}\)
\(\Rightarrow3S=1+\frac{2}{3}+.......+\frac{101}{3^{100}}\)
\(\Rightarrow3S-S=\left(1+\frac{2}{3}+..+\frac{101}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+..+\frac{101}{3^{101}}\right)\)
\(\Rightarrow2S=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{100}}-\frac{101}{3^{101}}< 1+\frac{1}{3}+....+\frac{1}{3^{100}}\)
\(\Rightarrow6S< 3+1+........+\frac{1}{3^{99}}\)
\(\Rightarrow6S-2S< \left(3+1+....+\frac{1}{3^{99}}\right)-\left(1+\frac{1}{3}+....+\frac{1}{3^{100}}\right)\)
\(\Rightarrow4S< 3-\frac{1}{3^{100}}< 3\Rightarrow S< \frac{3}{4}\)
S
18 tháng 3 2018
Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}+\frac{101}{3^{100}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{101}{3^{100}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+...+\frac{101}{3^{101}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}\)
\(6A=3+1+\frac{1}{3}+...+\frac{1}{3^{99}}-\frac{101}{3^{100}}\)
\(6A-2A=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{99}}-\frac{101}{3^{100}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}\right)\)
\(4A=3-\frac{101}{3^{100}}-\frac{1}{3^{100}}+\frac{101}{3^{101}}\)
\(4A=3-\frac{303}{3^{101}}-\frac{3}{3^{101}}+\frac{100}{3^{101}}\)
\(4A=3-\frac{206}{3^{101}}< 3\)
=>\(4A< 3\)
\(\Rightarrow A< \frac{3}{4}\)
21 tháng 5 2016
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\(\frac{4}{12}< \frac{9}{12}Vậy\frac{1}{3}< \frac{3}{4}\)
27 tháng 7 2019
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 \(F=1+\frac{1}{3}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3F=3+1+...+\frac{1}{3^{97}}\)
\(\Rightarrow3F-F=\left(3+...+\frac{1}{3^{97}}\right)-\left(1+...+\frac{1}{3^{98}}\right)\)
\(\Rightarrow2F=3-\frac{1}{3^{98}}< 3\)
\(\Rightarrow F< \frac{3}{2}\)
\(\Rightarrow2A< 4+\frac{3}{2}\)
\(\Rightarrow2A< \frac{11}{2}\)
\(\Rightarrow A< \frac{11}{4}\left(đpcm\right)\)
27 tháng 7 2019
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 )
\(\Rightarrow2D< 6\)
\(\Rightarrow D< 3\)
\(\Rightarrow2B< 11+3\)
\(\Rightarrow2B< 14\)
\(\Rightarrow B< 7\left(đpcm\right)\)
