Question

Tính:\(A=\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}+...+\dfrac{1}{210}\)

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Á huhu

Answer 1

Ta có:

\(A=\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{1}{210}\)

=> \(\dfrac{1}{2}A=\dfrac{1}{2}\left(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{1}{210}\right)\text{​}\)

\(=\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+...+\dfrac{1}{420}\)

\(=\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+...+\dfrac{1}{20.21}\)

\(=\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{20}-\dfrac{1}{21}\)

\(=\dfrac{1}{6}-\dfrac{1}{21}\)

\(=\dfrac{5}{42}\)

Vậy \(A=\dfrac{5}{42}\)