Question

$\frac{1}{1*3} +\frac{1}{3*5} +\frac{1}{5*7} +... +\frac{1}{99*101}$ Найти сумму

Answer 1

$\frac{1}{1\cdot3} +\frac{1}{3\cdot5} +\frac{1}{5\cdot7} +...+\frac{1}{99\cdot101} =\\=\frac{1}{2} (\frac{2}{1\cdot3} +\frac{2}{3\cdot5} +\frac{2}{5\cdot7} +...+\frac{2}{99\cdot101}) =\\=\frac{1}{2} (\frac{3-1}{1\cdot3} +\frac{5-3}{3\cdot5} +\frac{7-5}{5\cdot7} +...+\frac{101-99}{99\cdot101}) =\\=\frac{1}{2} (1-\frac{1}{3} +\frac{1}{3} - \frac{1}{5} +\frac{1}{5} - \frac{1}{7} +...+\frac{1}{99} - \frac{1}{101}) = \\=\frac{1}{2} (1-\frac{1}{101}) = \frac{1}{2} \cdot \frac{100}{101} = \frac{50}{101}$