CMR:\(\frac{1}{3}\)+\(\frac{2}{3^2}\)+\(\frac{3}{3^3}\)+...+\(\frac{100}{3^{100}}\)<\(\frac{3}{4}\)
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29 tháng 8 2016
hổng khó, marivan2016(mk bít nick thiệt nhưng hổng nói) làm ơn k giùm mk nha cảm ơn nhìu!!!
23 tháng 8 2016
\(3C=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+....+\frac{100}{3^{99}}.\)
\(2C=3C-C=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}-\frac{100}{3^{100}}.\)
\(2C=1+A-\frac{100}{3^{100}}\)
\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}=\frac{1}{2}\left(1-\frac{1}{3^{99}}\right)< \frac{1}{2}\)
=>\(2C=1+A-\frac{100}{3^{100}}< 1+\frac{1}{2}=\frac{3}{2}\)
\(C< \frac{3}{4}.\)
L
CMR:
\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+.....+\frac{100}{3^{100}}< \frac{3}{4}\)
4
1 tháng 1 2019
đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+....+\frac{100}{3^{100}}\)
\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6A=3+1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow4A=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}=3-\frac{203}{3^{100}}\)
\(\Rightarrow A=\frac{3-\frac{203}{3^{100}}}{4}\)
\(\Rightarrow A=\frac{3}{4}-\frac{203}{\frac{3^{100}}{4}}\le\frac{3}{4}\left(ĐPCM\right)\)
NH
1 tháng 1 2019
\(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)
\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow4D=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}=3-\frac{203}{3^{100}}\)
\(\Rightarrow D=\frac{3-\frac{203}{3^{100}}}{4}=\frac{3}{4}-\frac{203}{3^{100}.4}< \frac{3}{4}\left(đpcm\right)\)
Vậy \(D< \frac{3}{4}\)
Nguồn: @Dekisugi Hidetoshi
27 tháng 6 2020
Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+....\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow4A=A+3A=\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+.....\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+....\frac{1}{3^{98}}-\frac{1}{3^{99}}\Rightarrow4A< B\left(1\right)\)
\(\Rightarrow3B=3-1+\frac{1}{3}-\frac{1}{3^2}+....\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(4B=B+3B=3-\frac{1}{3^{99}}< 3\Rightarrow4B< 3\Rightarrow B< \frac{3}{4}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow4A< B< \frac{3}{4}\Rightarrow4A< \frac{3}{4}\Rightarrow A< \frac{3}{4}:4\Rightarrow A< \frac{3}{4}.\frac{1}{4}\Rightarrow A< \frac{3}{16}\)
=> đpcm.
BT
1
3 tháng 7 2016
\(ĐặtA=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6A-2A=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{203}{3^{100}}< 3\)
\(=>A< \frac{3}{4}\left(đpcm\right)\)
Ủng hộ mk nha ^_-
S
17 tháng 1 2018
Đặt \(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(3A+A=4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow4A< 1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\) (1)
Đặt \(B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(3B=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(B+3B=4B=3-\frac{1}{3^{98}}< 3\)
\(\Rightarrow B< \frac{3}{4}\) (2)
Từ (2) và (2) => \(4A< B< \frac{3}{4}\Rightarrow A< \frac{3}{16}\) (đpcm)
18 tháng 1 2018
\(A=\frac{7n-1}{4};B=\frac{5n+3}{12}\)
Tìm n để A,B đồng thời là các số nguyên tố
Ta có: \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(\frac{1}{3}A=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{100}{3^{101}}\)
\(\frac{2}{3}A=\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)-\left(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{100}{3^{101}}\right)\)
\(\frac{2}{3}A=\frac{1}{3}+\left(\frac{2}{3^2}-\frac{1}{3^2}\right)+\left(\frac{3}{3^3}-\frac{2}{3^3}\right)+...+\left(\frac{100}{3^{100}}-\frac{99}{3^{100}}\right)-\frac{100}{3^{101}}\)
\(=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{100}{3^{101}}\)
Đặt: \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(\Rightarrow\frac{1}{3}B=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{101}}\)
\(\Rightarrow\frac{2}{3}B=\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)-\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{101}}\right)\)
\(=\frac{1}{3}-\frac{1}{3^{101}}\)\(\Leftrightarrow B=\left(\frac{1}{3}-\frac{1}{3^{101}}\right):\frac{2}{3}=\left(\frac{1}{3}-\frac{1}{3^{101}}\right).\frac{3}{2}\)
Thay \(B\) vào \(\frac{2}{3}A\), ta có: \(\frac{2}{3}A=\frac{3}{2}\left(\frac{1}{3}-\frac{1}{3^{101}}\right)-\frac{100}{3^{101}}\)
\(\Rightarrow A=\left[\frac{3}{2}\left(\frac{1}{3}-\frac{1}{3^{101}}\right)-\frac{100}{3^{101}}\right]:\frac{2}{3}=\frac{9}{4}\left(\frac{1}{3}-\frac{1}{3^{101}}\right)-\frac{150}{3^{101}}\)
\(A=\frac{3}{4}-\frac{9}{4}.\frac{1}{3^{101}}-\frac{150}{3^{101}}\Rightarrow A< \frac{3}{4}\)
Vậy \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}< \frac{3}{4}\)(ĐPCM)
Xong.