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\(C=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+..+\frac{1}{2016}.\left(1+2+3+...+2016\right)\)
\(C=1+\frac{1}{2}.\left(1+2\right).2:2+\frac{1}{3}.\left(1+3\right).3:2+\frac{1}{4}.\left(1+4\right).4:2+...+\frac{1}{2016}.\left(1+2016\right).2016:2\)
\(C=1+3:2+4:2+5:2+...+2017:2\)
\(C=2.\frac{1}{2}+3.\frac{1}{2}+4.\frac{1}{2}+5.\frac{1}{2}+...+2017.\frac{1}{2}\)
\(C=\frac{1}{2}.\left(2+3+4+5+...+2017\right)\)
\(C=\frac{1}{2}.\left(2+2017\right).2016:2\)
\(C=\frac{1}{2}.2019.2016.\frac{1}{2}\)
\(C=2019.504=1017576\)
\(\frac{10^{2016}+2^3}{9}=\frac{10^{2016}-1}{9}+\frac{2^3+1}{9}=\left(1+10+10^2+...+10^{2015}\right)+1\in N.\)
\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)...\left(\frac{1}{2016}-1\right)\left(\frac{1}{2017}-1\right)\)
\(=\frac{-1}{2}.\frac{-2}{3}...\frac{-2015}{2016}.\frac{-2016}{2017}\)
\(=\frac{1.2...2015.2016}{2.3...2016.2017}\) ( tử số có chẵn số hạng )
\(=\frac{1}{2017}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{102}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{101}{102}=\frac{1}{102}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}}=\frac{C}{D}\)
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 \(B=\frac{C}{D}=\frac{C}{2016C}=\frac{1}{2016}\)
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{102}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{101}{102}=\frac{1\cdot2\cdot3\cdot....\cdot101}{2\cdot3\cdot4\cdot....\cdot102}\)
\(A=\frac{1}{102}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\left(\frac{2015}{1}+1\right)+\left(\frac{2014}{2}+1\right)+...+\left(\frac{1}{2015}+1\right)+1}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\frac{2016}{1}+\frac{2016}{2}+...+\frac{2016}{2015}+\frac{2016}{2016}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{2016\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}=\frac{1}{2016}\)
\(A=\left(\frac{1}{1^2}-1\right)\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2015^2}-1\right)\left(\frac{1}{2016^2}-1\right)\)
\(=0.\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2015^2}-1\right)\left(\frac{1}{2016^2}-1\right)=0>-\frac{1}{2}\)
suy ra A>B
A \(=\frac{1}{2016}\)