\(A=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{107}-\dfrac{1}{111}=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{12}{37}\\ B=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+...+\dfrac{5}{700}\\ \Rightarrow\dfrac{3}{5}B=\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+...+\dfrac{3}{25\cdot28}\\ \Rightarrow\dfrac{3}{5}B=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}=\dfrac{1}{4}-\dfrac{1}{28}=\dfrac{3}{14}\\ \Rightarrow B=\dfrac{3}{14}\cdot\dfrac{5}{3}=\dfrac{5}{14}\)
\(C=\dfrac{1}{2^1}+\dfrac{1}{2^3}+\dfrac{1}{2^5}+\dfrac{1}{2^7}+\dfrac{1}{2^9}\\ \Rightarrow4C=2+\dfrac{1}{2^1}+\dfrac{1}{2^3}+\dfrac{1}{2^5}+\dfrac{1}{2^7}\\ \Rightarrow4C-C=2-\dfrac{1}{2^9}\\ \Rightarrow3C=2-\dfrac{1}{512}=\dfrac{1023}{512}\\ \Rightarrow C=\dfrac{1023}{512}\cdot\dfrac{1}{3}=\dfrac{341}{512}\)
\(D=\dfrac{5}{1\times5\times8}+\dfrac{5}{5\times8\times12}+...+\dfrac{5}{33\times36\times40}\\ \Rightarrow\dfrac{5}{7}D=\dfrac{7}{1\times5\times8}+\dfrac{7}{5\times8\times12}+...+\dfrac{7}{33\times36\times40}\\ \Rightarrow\dfrac{5}{7}D=\dfrac{1}{1\times5}-\dfrac{1}{5\times8}+\dfrac{1}{5\times8}-\dfrac{1}{8\times12}+...+\dfrac{1}{33\times36}-\dfrac{1}{36\times40}\\ \Rightarrow\dfrac{5}{7}D=\dfrac{1}{5}-\dfrac{1}{1440}=\dfrac{287}{1440}\\ \Rightarrow D=\dfrac{2009}{7200}\)
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