Đặt \(A=\dfrac{1^2}{1.3}+\dfrac{2^2}{3.5}+\dfrac{3^3}{5.7}+...+\dfrac{1006^2}{2011.2013}\)

\(\Rightarrow4A=\dfrac{4.1^2}{1.3}+\dfrac{4.2^2}{3.5}+\dfrac{4.3^3}{5.7}+...+\dfrac{4.1006^2}{2011.2013}\)

\(\Rightarrow4A=1006+\dfrac{1}{2}.\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{2011}-\dfrac{1}{2013}\right]\)

\(\Rightarrow A=\dfrac{1006+\dfrac{1}{2}.\left(1-\dfrac{1}{2013}\right)}{4}\)

\(\Rightarrow A=251,6249\)

\(B=\dfrac{2^{24}\cdot3^5-2^{24}\cdot3^4}{2^{24}\cdot3^5}+1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{301}-\dfrac{1}{303}\)

\(=\dfrac{2^{24}\cdot3^4\left(3-1\right)}{2^{24}\cdot3^5}+\dfrac{302}{303}\)

\(=\dfrac{2}{3}+\dfrac{302}{303}=\dfrac{202+302}{303}=\dfrac{504}{303}\)

=168/101

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: \(D=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)\)

\(D=2\cdot\left(1-\dfrac{1}{2009}\right)=2\cdot\dfrac{2008}{2009}=\dfrac{4016}{2009}\)

\(A=\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)

\(A=\dfrac{2.2.3.3.4.4.5.5.6.6}{1.3.2.4.3.5.4.6.5.7}\)

\(A=\dfrac{2.3.4.5.6}{1.2.3.4.5}.\dfrac{2.3.4.5.6}{3.4.5.6.7}\)

\(A=\dfrac{6}{1}.\dfrac{2}{7}=\dfrac{12}{7}\)

\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{9.11}\right)\)

\(B=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{100}{99}\)

\(B=\dfrac{4.9.16.100}{3.8.15.99}\)

\(B=\dfrac{2.2.3.3.4.4.10.10}{1.3.2.4.3.5.9.11}\)

\(B=\dfrac{2.3.4.10}{1.2.3.9}.\dfrac{2.3.4.10}{3.4.5.11}\)

\(B=10.\dfrac{2}{11}=\dfrac{20}{11}\)

`A=2/[1.3]+2/[3.5]+2/[5.7]+.....+2/[99.101]`

`A=1-1/3+1/3-1/5+1/5-1/7+......+1/99-1/101`

`A=1-1/101=101-1/101=100/101`

\(\dfrac{100}{101}\)

A=2/1.3+2/3.5+2/5.7+...+2/99.101
   = 1/1 - 1/3 +1/3 - 1/5 +.... +1/99 - 1/101
   = 1-1/101
   =101/101-1/101
   =100/101   

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

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

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

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

a: \(A=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2022\cdot2024}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2022}-\dfrac{1}{2024}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{1011}{2024}=\dfrac{1011}{4848}< \dfrac{1}{4}\)

b: \(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2013\cdot2015}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2013}-\dfrac{1}{2015}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2014}{2015}=\dfrac{1007}{2015}< \dfrac{1}{2}\)