Question

Tính tổng sau:

S = \(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+\(\dfrac{1}{21}\)+...+\(\dfrac{1}{300}\)

Answer 1

\(S=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{300}\)

\(S=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{600}\)

\(S=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{24.25}\)

\(S=2.(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25})\)

\(S=2.(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25})\)

\(S=2.(\dfrac{1}{4}-\dfrac{1}{25})\)

\(S=2.\dfrac{21}{100}\)

\(S=\dfrac{21}{50}\)

\(\Rightarrow S=\dfrac{21}{50}\)

Answer 2

S = \(\dfrac{1}{10}\) + \(\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{300}\)

S = \(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{600}\)

S = \(\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{24.25}\)

S = 2 (\(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25}\))

S = 2 (\(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\))

S = 2 . (\(\dfrac{1}{4}-\dfrac{1}{25}\))

S = 2 . \(\dfrac{21}{100}\)

=> S = \(\dfrac{21}{50}\)