a: \(A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}=1-\dfrac{1}{99}=\dfrac{98}{99}\)
b: \(B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)
d: \(C=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{90}\)
\(=2\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=2\cdot\dfrac{4}{10}=\dfrac{8}{10}=\dfrac{4}{5}\)
`@`\(A=\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{97\times99}\)
\(A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(A=1-\dfrac{1}{99}\)
\(A=\dfrac{98}{99}\)
`@`\(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{90}\)
\(B=\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{9\times10}\)
\(B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(B=\dfrac{1}{2}-\dfrac{1}{10}\)
\(B=\dfrac{4}{10}=\dfrac{2}{5}\)
`@`\(D=\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{29}{30}+\dfrac{41}{42}+...+\dfrac{89}{90}\)
\(D=1-\dfrac{1}{6}+1-\dfrac{1}{12}+1-\dfrac{1}{30}+1-\dfrac{1}{42}+...+1-\dfrac{1}{90}\)
\(D=9-\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{90}\right)\)
\(D=9-\dfrac{2}{5}\)
\(D=\dfrac{43}{5}\)
`@`\(C=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{45}\)
\(C=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{90}\)
\(C=2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{90}\right)\)
\(C=2.\dfrac{2}{5}=\dfrac{4}{5}\)
`@`\(E=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}+\dfrac{1}{256}+\dfrac{1}{512}\)
\(E=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}+\dfrac{1}{2^7}+\dfrac{1}{2^8}\)
\(2.E=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}+\dfrac{1}{2^7}\)
\(E=2E-E\)
\(E=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}+\dfrac{1}{2^7}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}+\dfrac{1}{2^7}+\dfrac{1}{2^8}\right)\)
\(E=1-\dfrac{1}{2^8}\)
\(E=\dfrac{2^8-1}{2^8}\)
\(E=\dfrac{511}{512}\)
