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\(D=\dfrac{\left(2!\right)^2}{1^2}+\dfrac{\left(2!\right)^2}{3^2}+\dfrac{\left(2!\right)^2}{5^2}+...+\dfrac{\left(2!\right)^2}{2015^2}\)
\(D=\left(2!\right)^2\left(\dfrac{1}{3^2}+\dfrac{1}{5^2}+...+\dfrac{1}{2015^2}\right)\)
Xét số hạng tổng quát dạng: \(\dfrac{1}{\left(2n+1\right)^2}\) với \(n\in N\ge1\)
Ta có: \(\left(2n+1\right)^2-2n\left(2n+1\right)=1>0\)
\(\Rightarrow\left(2n+1\right)^2>2n\left(2n+1\right)\Rightarrow\dfrac{1}{\left(2n+1\right)^2}< \dfrac{1}{2n\left(2n+1\right)}\)
Do đó: \(\left\{{}\begin{matrix}\dfrac{1}{3^2}< \dfrac{1}{2.4}\\\dfrac{1}{5^2}< \dfrac{1}{4.6}\\....\\\dfrac{1}{2015^2}< \dfrac{1}{2014.2016}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}...+\dfrac{1}{2015^2}< 1+\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{2014.2016}\)
\(\Leftrightarrow\dfrac{D}{\left(2!\right)^2}< 1+\dfrac{1}{2.4}+\dfrac{1}{4.6}+..+\dfrac{1}{2014.2016}\)
\(\Leftrightarrow D< 4\left(1+\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{2014.2016}\right)\)
\(\Leftrightarrow D< 4+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{1007.1008}\)
\(\Leftrightarrow D< 4+\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+...+\dfrac{1008-1007}{1007.1008}\)
\(\Leftrightarrow D< 4+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{...1}{1007}-\dfrac{1}{1008}\)
\(\Leftrightarrow D< 5-\dfrac{1}{1008}< 5< 6\)
\(D=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)....\left(1-\dfrac{1}{2015}\right)\)
\(D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2014}{2015}=\dfrac{1.2.3....2014}{2.3.4....2015}\)
\(D=\dfrac{1}{2015}\)
\(D=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2015}\right)\)
\(D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2014}{2015}=\dfrac{1.2.3...2014}{2.3.4...2015}\)
\(D=\dfrac{1}{2015}\)
\(A=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2014^2}\right)\)
\(A=\dfrac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2012\cdot2014\right)\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2013\cdot2013\right)\left(2014\cdot2014\right)}\)
\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot2012\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2014\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)}\)
\(A=\dfrac{1\cdot2015}{2014\cdot2}=\dfrac{2015}{4028}\)
Vì \(\dfrac{2015}{4028}>-\dfrac{1}{2}\) nên A > B
e: \(=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}+\dfrac{2}{3}\right)-\dfrac{47}{24}=2-\dfrac{47}{24}=\dfrac{1}{24}\)
f: \(=-8\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-4:\dfrac{13}{12}=\dfrac{-48}{13}\)
g: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}-\dfrac{8}{4}=\dfrac{4}{25}+\dfrac{55}{4}-2=\dfrac{1191}{100}\)
Lời giải:
Ta có: \(D=(2!)^2\left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....+\frac{1}{2015^2}\right)\)
Xét số hạng tổng quát dạng \(\frac{1}{(2n+1)^2}\) với \(n\in\mathbb{N}\ge 1\)
Ta có: \((2n+1)^2-2n(2n+2)=1>0\)
\(\Rightarrow (2n+1)^2> 2n(2n+2)\Rightarrow \frac{1}{(2n+1)^2}< \frac{1}{2n(2n+2)}\)
Do đó: \(\left\{\begin{matrix} \frac{1}{3^2}< \frac{1}{2.4}\\ \frac{1}{5^2}< \frac{1}{4.6}\\ .....\\ \frac{1}{2015^2}< \frac{1}{2014.2016}\end{matrix}\right.\)
\(\Rightarrow \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{2015^2}< 1+\frac{1}{2.4}+\frac{1}{4.6}+....+\frac{1}{2014.1016}\)
\(\Leftrightarrow \frac{D}{(2!)^2}< 1+\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+....+\frac{1}{2014.2016}\)
\(\Leftrightarrow D< 4\left(1+\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+....+\frac{1}{2014.2016}\right)\)
\(\Leftrightarrow D< 4+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1007.1008}\)
\(\Leftrightarrow D< 4+\frac{2-1}{1,2}+\frac{3-2}{2.3}+...+\frac{1008-1007}{1007.1008}\)
\(\Leftrightarrow D< 4+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{107}-\frac{1}{1008}\)
\(\Leftrightarrow D< 5-\frac{1}{1008}< 5< 6\)