Question

Bài 6 : Tính

P=\(\dfrac{5}{1.4}\)+ \(\dfrac{5}{4.7}\)+\(\dfrac{5}{7.10}\)+. . . +\(\dfrac{5}{2017.2020}\)

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Answer 1

Ta có : P = \(\dfrac{5}{1.4}+\dfrac{5}{4.7}+\dfrac{5}{7.11}+...+\dfrac{5}{2017.2020}\)

\(\Rightarrow P=5\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.11}+...+\dfrac{1}{2017.2020}\right)\)

\(\Rightarrow\) \(P=5.\dfrac{3}{3}\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.11}+...+\dfrac{1}{2017.2020}\right)\)

\(\Rightarrow P=5.\dfrac{1}{3}\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{2017}-\dfrac{1}{2020}\right)\)

\(\Rightarrow P=5.\dfrac{1}{3}\left(1-\dfrac{1}{2020}\right)\)

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Answer 2

P=\(\dfrac{5}{1.4}+\dfrac{5}{4.7}+\dfrac{5}{7.10}+...+\dfrac{5}{2017.2020}\)

P=\(1-\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{1}{7}-\dfrac{1}{7}+...+\dfrac{1}{2017}-\dfrac{1}{2020}\)

P=1-\(\dfrac{1}{2020}\)

\(\Rightarrow P=\dfrac{2019}{2020}\)