Question

Tính B = \(\dfrac{1}{6}+\dfrac{1}{24}+\dfrac{1}{60}+.......+\dfrac{1}{990}\)

Answer 1

Ta có:\(B=\dfrac{1}{6}+\dfrac{1}{24}+\dfrac{1}{60}+...+\dfrac{1}{990}\)

\(2B=\dfrac{2}{6}+\dfrac{2}{24}+\dfrac{2}{60}+...+\dfrac{2}{990}\)

\(2B=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\dfrac{2}{3\cdot4\cdot5}+...+\dfrac{2}{9\cdot10\cdot11}\)

\(2B=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+\dfrac{1}{3\cdot4}-\dfrac{1}{4\cdot5}+...+\dfrac{1}{9\cdot10}-\dfrac{1}{10\cdot11}\)

\(2B=\dfrac{1}{1\cdot2}-\dfrac{1}{10\cdot11}\)

\(2B=\dfrac{27}{55}\)

\(B=\dfrac{27}{55}:2\)

\(B=\dfrac{27}{110}\)