Question

A=4/3*5+4/5*7+4/7*9+...+4/99*101

Answer 1

\(A=\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{99.101}.\)

\(A=\frac{4}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(A=2.\left(\frac{1}{3}-\frac{1}{101}\right)=2\cdot\frac{98}{303}=\frac{196}{303}\)

Answer 2

\(A=\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+....+\frac{4}{99.101}.\)

\(=2.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{99.101}\right)\)

\(=2.\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+.....+\frac{101-99}{99.101}\right)\)

\(=2.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\right)\)

\(=2.\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(=2.\frac{98}{303}=\frac{196}{303}\)