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CT
1
5 tháng 11 2016
\(C=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)
\(3C=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(3C-C=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)
\(2C=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6C=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6C-2C=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4C=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4C=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4C=3-\frac{203}{3^{100}}< 3\)
\(\Rightarrow C< \frac{3}{4}\left(đpcm\right)\)
NT
1
D
28 tháng 10 2018
Ta có:
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)
\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6A=3+1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow4A=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}=3-\frac{203}{3^{100}}\)
\(\Rightarrow A=\frac{3-\frac{203}{3^{100}}}{4}=\frac{3}{4}-\frac{203}{3^{100}.4}< \frac{3}{4}\Rightarrowđpcm\)
Vậy \(A< \frac{3}{4}\)
PH
2
14 tháng 10 2017
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}+\frac{101}{3^{100}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}+\frac{101}{3^{100}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}-\frac{101}{3^{101}}\)
\(2A=1+\left(\frac{1-\frac{1}{3^{100}}}{2}\right)-\frac{101}{3^{101}}< 1+\frac{1}{2}=\frac{3}{2}\)
\(\Rightarrow A< \frac{3}{2}:2=\frac{3}{4}\)( đpcm )
NT
0
NT
0
NT
0
DM
1
13 tháng 6 2018
Đặt :
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+.......+\frac{100}{3^{100}}\)
\(\Leftrightarrow3A=\frac{2}{3}+\frac{3}{3^2}+........+\frac{100}{3^{99}}\)
\(\Leftrightarrow3A-A=\left(\frac{2}{3}+\frac{3}{3^2}+.....+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+....+\frac{99}{3^{99}}+\frac{100}{3^{100}}\right)\)
\(\Leftrightarrow2A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
Đặt \(H=\frac{1}{3}+\frac{1}{3^2}+......+\frac{1}{3^{99}}\) \(\Leftrightarrow2A=H-\frac{100}{3^{100}}\)
\(\Leftrightarrow3H=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Leftrightarrow3H-H=\left(1+\frac{1}{3}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)\)
\(\Leftrightarrow2H=1-\frac{1}{3^{99}}< 1\)
\(\Leftrightarrow H=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
\(\Leftrightarrow2A=\frac{1-\frac{1}{3^{99}}}{2}-\frac{100}{3^{100}}\)
\(\Leftrightarrow A=\frac{1-\frac{1}{3^{99}}}{4}-\frac{100}{2.3^{100}}\)
\(\Leftrightarrow A< \frac{3}{4}\left(đpcm\right)\)
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