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\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{102}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{101}{102}=\frac{1}{102}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}}=\frac{C}{D}\)
Ta có: \(D=\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}\)(có 2015 số hạng)
\(D=\left(\frac{2015}{1}+1\right)+\left(\frac{2014}{2}+1\right)+...+\left(\frac{1}{2015}+1\right)-2015\)
\(D=2016+\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}-2015\)
\(D=\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}+1=\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}+\frac{2016}{2016}\)
\(D=2016\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}\right)=2016C\)
Vậy \(B=\frac{C}{D}=\frac{C}{2016C}=\frac{1}{2016}\)
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{102}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{101}{102}=\frac{1\cdot2\cdot3\cdot....\cdot101}{2\cdot3\cdot4\cdot....\cdot102}\)
\(A=\frac{1}{102}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\left(\frac{2015}{1}+1\right)+\left(\frac{2014}{2}+1\right)+...+\left(\frac{1}{2015}+1\right)+1}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\frac{2016}{1}+\frac{2016}{2}+...+\frac{2016}{2015}+\frac{2016}{2016}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{2016\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}=\frac{1}{2016}\)
mk nghỉ zậy nè ngọc ơi !!!
P=1+ 1/2 +1/3 +1/4 +....+1/100 < 0/1 +0/2 +0/3 +...+ 0/100
=>1+ 1/2 +1/3 +1/4 +....+1/100 < 0
=> 1+ 1/2 +1/3 +1/4 +....+1/100 \(\notin\)N
\(\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{2003}\right)\left(-1\frac{1}{2004}\right)\)
\(=-\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{2004}{2003}.\frac{2005}{2004}\)
\(=-\frac{3.4.5.....2004.2005}{2.3.4.....2003.2004}=\frac{-2005}{2}\)
