So sánh 1/101 + 1/102 + .......... + 1/199 + 1/200 và 5/8
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9 tháng 3 2017
Ta có :
\(\frac{1}{101}>\frac{1}{150}\)
\(\frac{1}{102}>\frac{1}{150}\)
\(\frac{1}{103}>\frac{1}{150}\)
\(..............\)
\(\frac{1}{150}=\frac{1}{150}\)
Cộng vế với vết ta được :
\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{150}>\frac{1}{150}+\frac{1}{150}+...+\frac{1}{150}\) (có 50 số hạng \(\frac{1}{150}\) ) \(=\frac{50}{150}=\frac{1}{3}\) \(\left(1\right)\)
Ta lại có :
\(\frac{1}{151}>\frac{1}{200}\)
\(\frac{1}{152}>\frac{1}{200}\)
\(\frac{1}{153}>\frac{1}{200}\)
\(............\)
\(\frac{1}{200}=\frac{1}{200}\)
Cộng vế với vết ta được :
\(\frac{1}{151}+\frac{1}{152}+...+\frac{1}{200}>\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\)(có 50 số hạng \(\frac{1}{200}\) ) \(=\frac{50}{200}=\frac{1}{4}\) \(\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+....+\frac{1}{200}>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
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Nguyễn Đức Trí
VIP
2 tháng 8 2023
\(A=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{120}\left(a\right)\)
\(\Rightarrow A=\left(\dfrac{1}{101}+\dfrac{1}{102}+...\dfrac{1}{125}\right)+\left(\dfrac{1}{126}+\dfrac{1}{127}+...\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...\dfrac{1}{175}\right)+\left(\dfrac{1}{176}+\dfrac{1}{177}+...\dfrac{1}{200}\right)\)
\(\Rightarrow A>25.\dfrac{1}{125}+25.\dfrac{1}{150}+25.\dfrac{1}{175}+25.\dfrac{1}{200}\)
\(\Rightarrow A>\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\)
\(\Rightarrow A>\dfrac{168+140+120+105}{840}=\dfrac{533}{840}>\dfrac{5}{8}\left(\dfrac{533}{840}>\dfrac{525}{840}\right)\)
\(\Rightarrow A>\dfrac{5}{8}\left(1\right)\)
\(\left(a\right)\Rightarrow A=\left(\dfrac{1}{101}+...\dfrac{1}{120}\right)+\left(\dfrac{1}{121}+...\dfrac{1}{140}\right)+\left(\dfrac{1}{141}+...\dfrac{1}{160}\right)+\left(\dfrac{1}{161}+...\dfrac{1}{180}\right)+\left(\dfrac{1}{181}+...\dfrac{1}{200}\right)\)
\(\Rightarrow A< 20.\dfrac{1}{100}+20.\dfrac{1}{120}+20.\dfrac{1}{140}+20.\dfrac{1}{160}+20.\dfrac{1}{180}\)
\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{504+420+360+315+280}{2520}=\dfrac{1879}{2520}< \dfrac{3}{4}\left(\dfrac{1879}{2520}< \dfrac{1890}{2520}\right)\)
\(\Rightarrow A< \dfrac{3}{4}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\dfrac{5}{8}< A< \dfrac{3}{4}\left(dpcm\right)\)
HH
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Ta đặt \(V=\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{199}+\dfrac{1}{200}\)
\(\Rightarrow V=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{125}\right)+\left(\dfrac{1}{126}+\dfrac{1}{127}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{175}\right)+\left(\dfrac{1}{176}+\dfrac{1}{177}+...+\dfrac{1}{200}\right)\)
\(\Rightarrow V>\left(\dfrac{1}{125}+\dfrac{1}{125}+...+\dfrac{1}{125}\right)+\left(\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{175}+\dfrac{1}{175}+...+\dfrac{1}{175}\right)+\left(\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}\right)\) (Mỗi nhóm có 25 số hạng.)
\(\Rightarrow V>\dfrac{1}{125}\times25+\dfrac{1}{150}\times25+\dfrac{1}{175}\times25+\dfrac{1}{200}\times25\)
\(\Rightarrow V>\dfrac{533}{840}>\dfrac{525}{840}=\dfrac{5}{8}\)
Vậy \(V>\dfrac{5}{8}\) hay \(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{5}{8}\)