\(A=\dfrac{1}{2}\left(2.\dfrac{2}{3}\right)\left(\dfrac{3}{2}.\dfrac{3}{4}\right)\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{2016}{2017}\)

\(A=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(\dfrac{1}{2015.2017}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right).....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)

\(=\dfrac{2016}{2017}\)

\(=\dfrac{1}{2}\cdot\dfrac{2^2-1+1}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2-1+1}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2016^2-1+1}{\left(2016-1\right)\left(2016+1\right)}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2016}{2015}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2016}{2017}\)

\(=\dfrac{1}{2}\cdot2016\cdot\dfrac{2}{2017}=\dfrac{2016}{2017}\)

\(A=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2014^2}\right)\)

\(A=\dfrac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2012\cdot2014\right)\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2013\cdot2013\right)\left(2014\cdot2014\right)}\)

\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot2012\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2014\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)}\)

\(A=\dfrac{1\cdot2015}{2014\cdot2}=\dfrac{2015}{4028}\)

Vì \(\dfrac{2015}{4028}>-\dfrac{1}{2}\) nên A > B

Ta có:
\(A=\dfrac{1}{2}.\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}...\dfrac{2016^2}{2015.2017}\)
\(A=\dfrac{1}{2}.\dfrac{2^2}{3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{2016^2}{2015.2017}\)
\(A=\left(\dfrac{2.3.4...2016}{2.3.4.5...2015}\right).\left(\dfrac{2.3.4...2016}{2.3.4.5...2017}\right)\)
\(A=2016.\dfrac{1}{2017}=\dfrac{2016}{2017}\)

Lời giải:

Xét tổng quát:

\(1+\frac{1}{k(k+2)}=\frac{k(k+2)+1}{k(k+2)}=\frac{(k+1)^2}{k(k+2)}\)

Thay $k=1,2,....,2015$ ta có:

\(1+\frac{1}{1.3}=\frac{2^2}{1.3}\)

\(1+\frac{1}{2.4}=\frac{3^2}{2.4}\)

\(1+\frac{1}{3.5}=\frac{4^2}{3.5}\)

\(1+\frac{1}{4.6}=\frac{5^2}{4.6}\)

.............

\(1+\frac{1}{2015.2017}=\frac{2016^2}{2015.2017}\)

Nhân theo vế:

\(\Rightarrow A=\frac{1}{2}\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{2015.2017}\right)\)

\(=\frac{1}{2}.\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}....\frac{2016^2}{2015.2017}\)

\(=\frac{(1.2.3...2016)^2}{(1.2.3...2015)(2.3.4...2017)}=\frac{(1.2.3...2016)(2.3....2016)}{(1.2.3...2015)(2.3.4...2017)}=2016.\frac{1}{2017}=\frac{2016}{2017}\)

\(A=\dfrac{1}{2}\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{2015.2017}\right)\)\(A=\dfrac{1}{2}.\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}...\dfrac{4064256}{4064255}\)

\(A=\dfrac{1}{2}.\dfrac{2.2.3.3.4.4....2016.2016}{3.8.15....4064255}\)

\(A=\dfrac{1}{2}.\dfrac{2.2.3.3.4.4....2016.2016}{1.3.2.4.3.5.....2015.2017}\)

\(A=\dfrac{1}{2}.\dfrac{2.3.4....2016}{1.2.3.....2015}.\dfrac{2.3.4.....2016}{3.4.5.....2017}\)

\(A=\dfrac{1}{2}.2016.\dfrac{2}{2017}\)

\(A=1008.\dfrac{2}{2017}\)

\(A=\dfrac{2016}{2017}\)