Bài 1:
c) \(A=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{99.101}\right)\)
\(=\dfrac{\left(1+3\right)}{1.3}.\dfrac{\left(1+8\right)}{2.4}.\dfrac{\left(1+15\right)}{3.5}...\dfrac{\left(1+9999\right)}{99.101}\)
\(=\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}...\dfrac{10000}{99.101}\)
\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}...\dfrac{100.100}{99.101}\)
\(=\left(\dfrac{2.3.4...99}{2.3.4...99}\right).\left(\dfrac{2.3.4...100^2}{3.4.5...99.101}\right)\)
\(=\dfrac{2.100^2}{101}=\dfrac{20000}{101}\)
d) \(D=\dfrac{2.2012}{1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+4+...+2012}}\)
\(=\dfrac{2.2012}{1+\dfrac{1}{\dfrac{2.3}{2}}+\dfrac{1}{\dfrac{3.4}{2}}+\dfrac{1}{\dfrac{4.5}{2}}+...+\dfrac{1}{\dfrac{2012.2013}{2}}}\)
\(=\dfrac{2.2012}{1+\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{2012.2013}}\)
\(=\dfrac{2.2012}{2.\left(\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2012.2013}\right)}\)
\(=\dfrac{2012}{\text{}\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2012}-\dfrac{1}{2013}}\)
\(=\dfrac{2012}{1-\dfrac{1}{2013}}\)
\(=\dfrac{2012}{\dfrac{2012}{2013}}=2013\)
