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Ta có: \(A=124\cdot\frac{1}{1984}\cdot\left(1-\frac{1}{1985}+\frac{1}{2}-\frac{1}{1986}+\frac{1}{3}-\frac{1}{1987}+...+\frac{1}{16}-\frac{1}{2000}\right)\)
\(\Rightarrow A=\frac{1}{16}\cdot\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+\frac{1}{1987}+...+\frac{1}{2000}\right)\right]\)
Laji cos: \(B=\frac{1}{16}\cdot\left(1-\frac{1}{17}+\frac{1}{2}-\frac{1}{18}+\frac{1}{3}-\frac{1}{19}+...+\frac{1}{1984}-\frac{1}{2000}\right)\)
\(\Rightarrow B=\frac{1}{16}\cdot\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1984}-\frac{1}{17}-\frac{1}{18}-\frac{1}{19}-...-\frac{1}{2000}\right)\)
\(\Rightarrow B=\frac{1}{16}\cdot\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1984}\right)-\left(\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+...+\frac{1}{2000}\right)\right]\)
Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}....\frac{18}{19}.\frac{19}{20}\)
\(=\frac{1.2....18.19}{2.3...19.20}\)
\(=\frac{1}{20}>\frac{1}{21}\)
Vậy A > 1/21
Lời giải:
\(16B=\frac{17-1}{1.17}+\frac{18-2}{2.18}+...+\frac{2012-1996}{1996.2012}\\ =1-\frac{1}{17}+\frac{1}{2}-\frac{1}{18}+...+\frac{1}{1996}-\frac{1}{2012}\\ =(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1996})-(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{2012})\\ =(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16})-(\frac{1}{1997}+...+\frac{1}{2012})\\ B=\frac{1}{16}[(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16})-(\frac{1}{1997}+...+\frac{1}{2012})]\)
Lại có:
$1984A=124(\frac{1985-1}{1.1985}+\frac{1986-2}{2.1986}+...+\frac{2012-28}{28.2012})$
$=124(1-\frac{1}{1985}+\frac{1}{2}-\frac{1}{1986}+\frac{1}{3}-\frac{1}{1987}+....+\frac{1}{28}-\frac{1}{2012})$
$=124[(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{28})-(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2012})]$
$A=\frac{124}{1984}[(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{28})-(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2012})]$
$=\frac{1}{16}[(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{28})-(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2012})]$
$=\frac{1}{16}[(1+\frac{1}{2}+...+\frac{1}{16})+(\frac{1}{17}-\frac{1}{1985})+(\frac{1}{18}-\frac{1}{1986})+...+(\frac{1}{28}-\frac{1}{1996})-(\frac{1}{1997}+\frac{1}{1998}+...+\frac{1}{2012})]$
$> \frac{1}{16}[(1+\frac{1}{2}+...+\frac{1}{16})-(\frac{1}{1997}+\frac{1}{1998}+...+\frac{1}{2012})]=B$
Vậy $A>B$