Question

A=\(\left(\frac{1}{8}+\frac{1}{8\cdot15}+\frac{1}{15\cdot22}+..+\frac{1}{43\cdot50}\right)\cdot\frac{4-3-5-7-...-49}{217}\)

Answer 1

A = \(\left(\frac{1}{8}+\frac{1}{8.15}+\frac{1}{15.22}+...+\frac{1}{43.50}\right)\cdot\frac{4-3-5-7-...-49}{217}\)

A = \(\frac{1}{7}.\left(\frac{7}{1.8}+\frac{7}{8.15}+\frac{7}{15.22}+...+\frac{7}{43.50}\right)\cdot\frac{4-\left(3+5+7+...+49\right)}{217}\)

A = \(\frac{1}{7}.\left(1-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+\frac{1}{15}-\frac{1}{22}+...+\frac{1}{43}-\frac{1}{50}\right)\cdot\frac{4-\left(49+3\right)\left[\left(49-3\right):2+1\right]:2}{217}\)

A = \(\frac{1}{7}\cdot\left(1-\frac{1}{50}\right)\cdot\frac{4-52.24:2}{217}\)

A = \(\frac{1}{7}\cdot\frac{49}{50}\cdot\frac{4-624}{217}\)

A = \(\frac{7}{50}\cdot\frac{-620}{217}=-\frac{2}{5}\)