Решить 1/6*7+1/7*8+...+ 1/49*50=

$\frac{1}{6*7}+\frac{1}{7*8}+...+\frac{1}{49*50}=\\\\\frac{7-6}{6*7}+\frac{8-7}{7*8}+...+\frac{50-49}{49*50}=\\\\\frac{7}{6*7}-\frac{6}{6*7}+\frac{8}{7*8}-\frac{7}{7*8}+...+\frac{50}{49*50}-\frac{49}{49*50}=\\\\\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{49}-\frac{1}{50}=\\\\\frac{1}{6}-\frac{1}{50}=\frac{1*25}{6*25}-\frac{1*3}{50*3}=\frac{25-3}{150}=\frac{22}{150}=\frac{11}{75}$
ответ: 11/75