Question

(\(\dfrac{1}{8}+\dfrac{1}{8.15}+\dfrac{1}{15.22}+....+\dfrac{1}{43.50}\)) . \(\dfrac{4-3-5-7-...-49}{217}\)

Answer 1

Chỗ phức tạp là ở biểu thức trong ngoặc thôi

Ta có

\(\dfrac{1}{8}+\dfrac{1}{8\cdot15}+\dfrac{1}{15\cdot22}...+\dfrac{1}{43\cdot50}\)

\(=\dfrac{1}{8}\cdot\left[\dfrac{1}{7}\left(\dfrac{1}{8}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{22}+....+\dfrac{1}{43}-\dfrac{1}{50}\right)\right]\)

\(=\dfrac{1}{8}\cdot\left[\dfrac{1}{7}\left(\dfrac{1}{8}-\dfrac{1}{50}\right)\right]=\dfrac{1}{8}\cdot\dfrac{3}{200}=\dfrac{3}{1600}\)