Violympic toán 6

Nguyễn Ngọc Thảo

find the value of:

\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+\(\dfrac{1}{21}\)+...+\(\dfrac{1}{300}\)

Huy Thắng Nguyễn
6 tháng 9 2017 lúc 18:43

Đặt \(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{300}\)

\(=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{600}\)

\(=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{24.25}\)

\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{24}-\dfrac{1}{25}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{25}\right)\)

\(=2\cdot\dfrac{23}{50}=\dfrac{23}{25}\)

Vậy A = \(\dfrac{23}{25}\).

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Nam Nguyễn
6 tháng 9 2017 lúc 18:53

It's very easy :)))))

Call expression is A, we have:

\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{300}.\)

\(\Rightarrow2A=2\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{300}\right).\)

\(\Rightarrow2A=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{600}.\)

\(\Rightarrow2A=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{24.25}.\)

\(\Rightarrow2A=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\right).\)

\(\Rightarrow2A=2\left(\dfrac{1}{2}-\dfrac{1}{25}\right).\)

\(\Rightarrow2A=2\left(\dfrac{25}{50}-\dfrac{2}{50}\right).\)

\(\Rightarrow2A=2.\dfrac{23}{50}=\dfrac{23}{25}.\)

Vậy.....

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