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\(S=\dfrac{1}{2}.\left(\dfrac{2}{\sqrt{1.3}}+\dfrac{2}{\sqrt{3.5}}+.......+\dfrac{2}{\sqrt{29.31}}\right)\)
\(S=\dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}+\dfrac{1}{\sqrt{5}}+.....-\dfrac{1}{\sqrt{29}}+\dfrac{1}{\sqrt{29}}-\dfrac{1}{\sqrt{31}}\right)\)
\(S=\dfrac{1}{2}.\left(1-\dfrac{1}{\sqrt{31}}\right)=\dfrac{1}{2}.\left(\dfrac{31-\sqrt{31}}{31}\right)=\dfrac{31-\sqrt{31}}{62}\)
\(=\dfrac{1}{2}\cdot\dfrac{2^2-1+1}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2-1+1}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2016^2-1+1}{\left(2016-1\right)\left(2016+1\right)}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2016}{2015}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2016}{2017}\)
\(=\dfrac{1}{2}\cdot2016\cdot\dfrac{2}{2017}=\dfrac{2016}{2017}\)
\(A=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(\dfrac{1}{2015.2017}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right).....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{2016}{2017}\)
\(A=\dfrac{1}{2}\left(\dfrac{2.2}{1.3}\right).\left(\dfrac{3.3}{2.4}\right)...\left(\dfrac{2020.2020}{2019.2021}\right)\)
\(=\dfrac{1.2.2.3.3...2020.2020}{1.2.2.3.3.4.4...2019.2021}\)
\(=\dfrac{1}{2021}\)
\(A=\dfrac{1}{2}\cdot\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\left(1+\dfrac{1}{3\cdot5}\right)...\left(1+\dfrac{1}{2019\cdot2021}\right)\)
\(A=\dfrac{1}{2}\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\left(1+\dfrac{1}{4^2-1}\right)...\left(1+\dfrac{1}{2020^2-1}\right)\)
\(A=\dfrac{1}{2}\cdot\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\cdot\left(3+1\right)}...\left(\dfrac{2020^2}{\left(2020-1\right)\cdot\left(2020+1\right)}\right)\)
\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{2}{3}\cdot\dfrac{3}{2}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2020}{2021}\)
\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}\)
\(A=\dfrac{1}{2}\cdot2020\cdot\dfrac{2}{2021}=\dfrac{2020}{2021}\)
2S=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2007.2009}\)
=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{2007}-\dfrac{1}{2009}\)
= 1- \(\dfrac{1}{2009}\)
= \(\dfrac{2008}{2009}\)
=> S=\(\dfrac{1004}{2009}\)
\(A=\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)
\(A=\dfrac{2.2.3.3.4.4.5.5.6.6}{1.3.2.4.3.5.4.6.5.7}\)
\(A=\dfrac{2.3.4.5.6}{1.2.3.4.5}.\dfrac{2.3.4.5.6}{3.4.5.6.7}\)
\(A=\dfrac{6}{1}.\dfrac{2}{7}=\dfrac{12}{7}\)
\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{9.11}\right)\)
\(B=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{100}{99}\)
\(B=\dfrac{4.9.16.100}{3.8.15.99}\)
\(B=\dfrac{2.2.3.3.4.4.10.10}{1.3.2.4.3.5.9.11}\)
\(B=\dfrac{2.3.4.10}{1.2.3.9}.\dfrac{2.3.4.10}{3.4.5.11}\)
\(B=10.\dfrac{2}{11}=\dfrac{20}{11}\)
I: Để 3n+4/n+2 là số nguyên thì \(3n+4⋮n+2\)
\(\Leftrightarrow3n+6-2⋮n+2\)
\(\Leftrightarrow n+2\in\left\{1;-1;2;-2\right\}\)
hay \(n\in\left\{-1;-3;0;-4\right\}\)
II: \(D=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)\)
\(D=2\cdot\left(1-\dfrac{1}{2009}\right)=2\cdot\dfrac{2008}{2009}=\dfrac{4016}{2009}\)