Question

a)    B = \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.......+\frac{4}{99.101}\)

b)    C = \(\frac{5}{1.5}+\frac{5}{5.9}+\frac{5}{9.13}+......+\frac{5}{101.105}\)

Answer 1

a, \(\frac{1}{2}.B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

      \(\frac{1}{2}.B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

         \(\frac{1}{2}.B=1-\frac{1}{101}=\frac{100}{101}\)

                  \(B=\frac{100}{101}.2=\frac{200}{101}\)

b, \(\frac{4}{5}.C=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{101.105}\)

      \(\frac{4}{5}.C=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)

          \(\frac{4}{5}.C=1-\frac{1}{105}=\frac{104}{105}\)

                 \(C=\frac{104}{105}.\frac{5}{4}=\frac{26}{21}\)

Answer 2

\(B=\frac{2}{2}\cdot\left(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+....+\frac{4}{99\cdot101}\right)\)

\(=\frac{4}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right)\)

\(=2\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=2\cdot\left(1-\frac{1}{101}\right)\)

\(=2\cdot\frac{100}{101}\)

\(=1\frac{99}{101}\)

Answer 3

a) \(B=\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{99\cdot101}\)

\(\Rightarrow\frac{2}{4}B=\frac{2}{4}\left(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{99\cdot101}\right)\)

\(\Leftrightarrow\frac{2}{4}B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{99\cdot101}\)

\(\Leftrightarrow\frac{2}{4}B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{101}\)

\(\Leftrightarrow\frac{2}{4}B=1-\frac{1}{101}=\frac{100}{101}\)

\(\Leftrightarrow B=\frac{100}{101}:\frac{2}{4}=\frac{100\cdot4}{101\cdot2}=\frac{200}{101}\)

Answer 4

b) \(C=\frac{5}{1\cdot5}+\frac{5}{5\cdot9}+\frac{5}{9\cdot13}+...+\frac{5}{101\cdot105}\)

\(\Rightarrow\frac{4}{5}C=\frac{4}{5}\left(\frac{5}{1\cdot5}+\frac{5}{5\cdot9}+\frac{5}{9\cdot13}+....+\frac{5}{101\cdot105}\right)\)

\(\Leftrightarrow\frac{4}{5}C=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+....+\frac{4}{101\cdot105}\)

\(\Leftrightarrow\frac{4}{5}C=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+....+\frac{1}{101}-\frac{1}{105}\)

\(\Leftrightarrow\frac{4}{5}C=1-\frac{1}{105}=\frac{104}{105}\)

\(\Leftrightarrow C=\frac{104}{105}:\frac{4}{5}=\frac{104\cdot5}{105\cdot4}=\frac{26}{21}\)