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B=\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{20}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{19}{20}\)
\(=\dfrac{1.2.3....19}{2.3.4.....20}\)
\(=\dfrac{1.2.3....19:\left(2.3.....19\right)}{2.3.4.....20:\left(2.3.4.....19\right)}\)
\(=\dfrac{1}{20}\)
\(-\frac{1}{2}+\frac{1}{3}+\left(-\frac{1}{4}\right)+\left(-\frac{2}{8}\right)+\frac{4}{18}+\frac{4}{9}\)
= \(0\)
a: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{100\cdot101}\)
=1-1/2+1/2-1/3+...+1/100-1/101
=1-1/101=100/101
b: \(A=1+\dfrac{1}{2}+1+\dfrac{1}{6}+1+\dfrac{1}{12}+...+1+\dfrac{1}{10100}\)
\(=100+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)\)
\(=101-\dfrac{1}{101}< 101\)
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)=\frac{-3}{2^2}.\frac{-8}{3^2}...\frac{-9999}{100^2}\)
\(=-\frac{3.8...9999}{2^2.3^2...100^2}=-\frac{1.3.2.4...99.101}{2.2.3.3...100.100}=-\frac{\left(1.2....99\right).\left(3.4...101\right)}{\left(2.3...100\right).\left(2.3...100\right)}=-\frac{1.101}{100.2}=-\frac{101}{200}\)
\(< -\frac{100}{200}=\frac{1}{2}=B\)
=> A < B
\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2012}}\)
\(2A=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2011}}\)
\(2A-A=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2011}}-1-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{2012}}\)
\(A=2-\dfrac{1}{2^{2012}}\)