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M=1/10 + 1/15 + 1/21 +....+ 1/120
M=2/20 +2/30+2/42+....+2/240
M=2/4.5 + 2/5.6 + 2/6.7 +.....+ 2/15.16
M=2.(1/4.5 +......+ 1/15.16)
M=2.(1/4 -1/5 +1/5 - 1/6 +.....+ 1/15 - 1/16)
M=2.(1/4 - 1/16)
M=2.(4/16 - 1/16)
M=2. 3/16
M=6/16=3/8
Có 1/3 = 8/24 < 9/24 = 3/8 =>1/3<M
Có 1/2 = 4/8>3/8 =>1/2 >M
=> 1/3 < M < 1/2
\(M=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{105}+\frac{1}{120}\)
\(M=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+...+\frac{2}{210}+\frac{2}{240}\)
\(M=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+...+\frac{2}{14.15}+\frac{2}{15.16}\)
\(M=\frac{2}{4}-\frac{2}{5}+\frac{2}{5}-\frac{2}{6}+\frac{2}{6}-\frac{2}{7}+\frac{2}{7}-\frac{2}{8}+...+\frac{2}{15}-\frac{2}{16}\)
\(M=\frac{2}{4}-\frac{2}{16}=\frac{3}{8}\)
Vì \(\frac{3}{9}< \frac{3}{8}< \frac{4}{8}\)nên \(\frac{1}{3}< M< \frac{1}{2}\)
Vậy \(\frac{1}{3}< M< \frac{1}{2}\)
P/S : Đừng nói như lần trước nhé!
\(M=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{105}+\dfrac{1}{120}\)
\(M=2.\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)
\(M=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{15.16}\right)\)
\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(M=2.\dfrac{3}{16}\)
\(M=\dfrac{3}{8}\)
Vậy \(\dfrac{1}{3}< M< \dfrac{1}{2}\)
= 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 +...+ 1/14 - 1/15 + 1/15 - 1/16
= 1/4 - 1/16
= 3/16
Công thức : 1/ a x (a+1) = 1/a - 1/ a+1
\(A = (\frac{1}{10} + ...+ \frac{1}{19} ) + (\frac{1}{20} + ...+ \frac{1}{29}) + (\frac{1}{30} +...+ \frac{1}{39} ) + (\frac{1}{40} + ...+\frac{1}{49} ) + (\frac{1}{50} +....+ \frac{1}{59}) + (\frac{1}{60} + ....+\frac{1}{69}) + \frac{1}{70}\)
Ta có : mỗi bên có 10 số hạng
\( (\frac{1}{10} + ..+ \frac{1}{19}) < (\frac{1}{10} + ...+ \frac{1}{10}) = \frac{1}{1}\)
\(\frac{1}{20}+..+ \frac{1}{29} < (\frac{1}{20}+..+\frac{1}{20}) = \frac{1}{2}\)
\((\frac{1}{30} +...+ \frac{1}{39} )< (\frac{1}{30} +...+ \frac{1}{30}) = \frac{1}{3}\)
\((\frac{1}{40} + ...+\frac{1}{49} )< (\frac{1}{40} + ...+\frac{1}{40}) = \frac{1}{4}\)
\((\frac{1}{50} +....+ \frac{1}{59})< (\frac{1}{50} +....+ \frac{1}{50}) = \frac{1}{5}\)
\((\frac{1}{60} + ....+\frac{1}{69}) + \frac{1}{70}< (\frac{1}{60} + ....+\frac{1}{60})+ \frac{1}{70} = \frac{1}{6} +\frac{1}{70}\)
\(\implies A < 1+\frac{1}{2} + ...+ \frac{1}{6} + \frac{1}{70}= \frac{13}{15} + \frac{1}{70} <1<\frac {51}{20} \)
\(\implies A<\frac{51}{20}\) \((đpcm)\)