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Ta có:
\(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{150}>\dfrac{1}{150}+\dfrac{1}{150}+\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\) (có 50 số hạng)
⇔ \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{150}>\dfrac{1}{3}\) \(\left(1\right)\)
\(\dfrac{1}{151}+\dfrac{1}{152}+\dfrac{1}{153}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}\) (có 50 số hạng)
⇔ \(\dfrac{1}{151}+\dfrac{1}{152}+\dfrac{1}{153}+...+\dfrac{1}{200}>\dfrac{1}{4}\) \(\left(2\right)\)
Từ (1) và (2), cộng vế theo vế. Ta được:
\(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{150}+\dfrac{1}{151}+\dfrac{1}{152}+\dfrac{1}{153}+...+\dfrac{1}{200}>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\)
⇒ \(ĐPCM\)
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{199}-\dfrac{1}{200}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{199}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+..+\dfrac{1}{200}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{200}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)\)
\(=\dfrac{1}{101}+...+\dfrac{1}{199}+\dfrac{1}{200}\)
bạn ơi cái câu <1 số hạng cuối cùng là j thế?
Ta có:
\(\dfrac{1}{101}>\dfrac{1}{150}\)
\(\dfrac{1}{102}>\dfrac{1}{150}\)
....
\(\dfrac{1}{150}=\dfrac{1}{150}\)
=>\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}>\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\)(50 số)=\(\dfrac{1}{3}\)
Ta có:
\(\dfrac{1}{152}>\dfrac{1}{200}\)
\(\dfrac{1}{153}>\dfrac{1}{200}\)
....
\(\dfrac{1}{200}=\dfrac{1}{200}\)
=>\(\dfrac{1}{151}+\dfrac{1}{153}+...+\dfrac{1}{120}>\dfrac{1}{120}+\dfrac{1}{120}+...+\dfrac{1}{120}\)(50 số)=\(\dfrac{1}{4}\)
=>\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}>\dfrac{1}{3}+\dfrac{1}{4}\)
=> \(A>\dfrac{7}{12}\)
a, Ta có :
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{1\cdot2\cdot3\cdot4}+...+\dfrac{1}{1\cdot2\cdot3\cdot...\cdot100}\\ < \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}< 1\\ \Rightarrow M< 1\\ \RightarrowĐpcm\)
Ta có: \(C=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)
\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{120}\right)+\left(\dfrac{1}{121}+\dfrac{1}{122}+\dfrac{1}{123}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+\dfrac{1}{153}+...+\dfrac{1}{180}\right)+\left(\dfrac{1}{181}+\dfrac{1}{182}+\dfrac{1}{183}+...+\dfrac{1}{200}\right)\)
\(\Leftrightarrow C>20\cdot\dfrac{1}{120}+30\cdot\dfrac{1}{150}+30\cdot\dfrac{1}{180}+20\cdot\dfrac{1}{200}\)
\(\Leftrightarrow C>\dfrac{1}{6}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{10}=\dfrac{19}{30}=\dfrac{76}{120}\)
\(\Leftrightarrow C>\dfrac{75}{120}=\dfrac{5}{8}\)
hay \(C>\dfrac{5}{8}\)(đpcm)
Ta có: \(\dfrac{3}{10}>\dfrac{3}{15}\)
\(\dfrac{3}{11}>\dfrac{3}{15}\)
\(\dfrac{3}{12}>\dfrac{3}{15}\)
\(\dfrac{3}{13}>\dfrac{3}{15}\)
\(\dfrac{3}{14}>\dfrac{3}{15}\)
Do đó: \(\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}>\dfrac{3}{15}+\dfrac{3}{15}+\dfrac{3}{15}+\dfrac{3}{15}+\dfrac{3}{15}=1\)
hay 1<S(1)
Ta có: \(\dfrac{3}{11}< \dfrac{3}{10}\)
\(\dfrac{3}{12}< \dfrac{3}{10}\)
\(\dfrac{3}{13}< \dfrac{3}{10}\)
\(\dfrac{3}{14}< \dfrac{3}{10}\)
Do đó: \(\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}< \dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}+\dfrac{3}{10}=\dfrac{12}{10}\)
\(\Leftrightarrow S< \dfrac{15}{10}=\dfrac{3}{2}< 2\)(2)
Từ (1) và (2) suy ra 1<S<2(đpcm)
\(A=\dfrac{\left(3+\dfrac{2}{15}+\dfrac{1}{5}\right):\dfrac{5}{2}}{\left(5+\dfrac{3}{7}-2-\dfrac{1}{4}\right):\left(4+\dfrac{43}{56}\right)}\)
\(=\dfrac{\dfrac{10}{3}\cdot\dfrac{2}{5}}{\dfrac{89}{28}:\dfrac{267}{56}}=\dfrac{4}{3}:\dfrac{2}{3}=2\)
\(B=\dfrac{\dfrac{6}{5}:\left(\dfrac{6}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{8}{25}+\dfrac{2}{25}}=\dfrac{\dfrac{6}{5}:\dfrac{3}{2}}{\dfrac{2}{5}}=2\)
Do đó: A=B
Chứng minh :
\(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{200}< \dfrac{25}{12}\)
Ta có :
\(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{200}\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{200}\right)\)
\(=\dfrac{25}{12}+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{200}\right)>\dfrac{25}{12}\)
Vậy 1 +12+13+...+1200>2512