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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=\(\frac{1}{3}\)

\(\frac{1}{3}và\frac{3}{4}\)

\(\frac{1}{3}=\frac{4}{12}\)

\(\frac{3}{4}=\frac{9}{12}\)

\(\frac{4}{12}< \frac{9}{12}Vậy\frac{1}{3}< \frac{3}{4}\)

18 tháng 3 2018

Đặt \(S=\frac{1}{3}+\frac{2}{3^2}+.......+\frac{101}{3^{101}}\)

\(\Rightarrow3S=1+\frac{2}{3}+.......+\frac{101}{3^{100}}\)

\(\Rightarrow3S-S=\left(1+\frac{2}{3}+..+\frac{101}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+..+\frac{101}{3^{101}}\right)\)

\(\Rightarrow2S=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{100}}-\frac{101}{3^{101}}< 1+\frac{1}{3}+....+\frac{1}{3^{100}}\)

\(\Rightarrow6S< 3+1+........+\frac{1}{3^{99}}\)

\(\Rightarrow6S-2S< \left(3+1+....+\frac{1}{3^{99}}\right)-\left(1+\frac{1}{3}+....+\frac{1}{3^{100}}\right)\)

\(\Rightarrow4S< 3-\frac{1}{3^{100}}< 3\Rightarrow S< \frac{3}{4}\)

18 tháng 3 2018

Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\)

\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}+\frac{101}{3^{100}}\)

\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{101}{3^{100}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+...+\frac{101}{3^{101}}\right)\)

\(2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}\)

\(6A=3+1+\frac{1}{3}+...+\frac{1}{3^{99}}-\frac{101}{3^{100}}\)

\(6A-2A=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{99}}-\frac{101}{3^{100}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}\right)\)

\(4A=3-\frac{101}{3^{100}}-\frac{1}{3^{100}}+\frac{101}{3^{101}}\)

\(4A=3-\frac{303}{3^{101}}-\frac{3}{3^{101}}+\frac{100}{3^{101}}\)

\(4A=3-\frac{206}{3^{101}}< 3\)

=>\(4A< 3\)

\(\Rightarrow A< \frac{3}{4}\)

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 )

\(\Rightarrow2D< 6\)

\(\Rightarrow D< 3\)

\(\Rightarrow2B< 11+3\)

\(\Rightarrow2B< 14\)

\(\Rightarrow B< 7\left(đpcm\right)\)

12 tháng 8 2016

Bài 1:

C = 1/101 + 1/102 + 1/103 + ... + 1/200

Có:

C < 1/101 + 1/101 + 1/101 + ... + 1/101

C < 100 . 1/101

C < 100/101

Mà 100/101 < 1

=> C < 1 (1)

Có:

C > 1/200 + 1/200 + 1/200 + ... + 1/200

C > 100 . 1/200

C > 1/2 (2)

Từ (1) và (2)

=> 1/2<C<1

Ủng hộ nha mk làm tiếp

29 tháng 3 2017

\(B=-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)

Suy ra \(3B=-1+\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-.....+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)

\(3B-B=-1-\frac{1}{3^{101}}\)hay \(2B=-1-\frac{1}{3^{101}}\)

Khi đó \(B=\frac{-1}{2}-\frac{1}{3^{101}.2}\)