\(D=\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^5+\left(\frac{1}{2}\right)^7+......+\left(\frac{1}{2}\right)^{2015}\)
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\(D=2!^2\left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{2015^2}\right)\)
tổng trong ngoặc nhỏ hơn 1 nên D nhỏ hơn 4.1=4<6
Vậy Đ<6
\(C=\frac{5}{2}\cdot\frac{7}{5}\cdot\frac{9}{7}\cdot\frac{11}{9}\cdot...\cdot\frac{2017}{2015}\cdot\frac{2019}{2017}=\frac{2019}{2}\)
\(D=\left(1-\frac{1}{\frac{2\cdot3}{2}}\right)\cdot\left(1-\frac{1}{\frac{3\cdot4}{2}}\right)\cdot\left(1-\frac{1}{\frac{4\cdot5}{2}}\right)\cdot\left(1-\frac{1}{\frac{5\cdot6}{2}}\right)\cdot...\cdot\left(1-\frac{1}{\frac{39\cdot40}{2}}\right)\)
\(=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot\left(1-\frac{2}{5\cdot6}\right)\cdot...\cdot\left(1-\frac{2}{39\cdot40}\right)\cdot\)
Nhận xét: \(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)nên:
\(D=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\frac{6\cdot3}{4\cdot5}\cdot\frac{7\cdot4}{5\cdot6}\cdot\frac{8\cdot5}{6\cdot7}\cdot...\cdot\frac{41\cdot38}{39\cdot40}=\)
\(D=\frac{4\cdot5\cdot6\cdot7\cdot...\cdot41\times1\cdot2\cdot3\cdot4\cdot...\cdot38}{2\cdot3\cdot4\cdot5\cdot...\cdot39\times3\cdot4\cdot5\cdot6\cdot..\cdot40}=\frac{1}{39}\cdot\frac{41}{3}=\frac{41}{117}\)
\(\left(1-\frac{2}{5}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{2}{2013}\right).\left(1-\frac{2}{2015}\right)\)
\(=\frac{3}{5}.\frac{5}{7}...\frac{2011}{2013}.\frac{2013}{2015}\)
\(=\frac{3.5...2011.2013}{5.7...2013.2015}\)
\(=\frac{3}{2015}\)