Question

A = \(\frac{1}{1\times4}\)+ \(\frac{1}{4\times7}\)+ . . . . + \(\frac{1}{2017\times2020}\)

Answer 1

\(3A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{2017\cdot2020}\)

\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2020}\)

\(3A=1-\frac{1}{2020}\)

\(A=\frac{673}{2020}\)

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Answer 2

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Answer 3

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Answer 4

\(A=\frac{1}{1×4}+\frac{1}{4×7}+...+\frac{1}{2017×2020}\text{ }\)

\(A=\frac{1×3}{\left(1×4\right)×3}+\frac{1×3}{\left(4×7\right)×3}+...+\frac{1×3}{\left(2017×2020\right)×3}\)

\(A=\frac{1}{3}×\left(\frac{3}{1×4}+\frac{3}{4×7}+...+\frac{3}{2017×2020}\right)\)

\(A=\frac{1}{3}×\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2020}\right)\)

\(A=\frac{1}{3}×\left(1-\frac{1}{2020}\right)\)

\(A=\frac{1}{3}×\left(\frac{2020}{2020}-\frac{1}{2020}\right)\)

\(A=\frac{1}{3}×\frac{2019}{2020}\)

\(A=\frac{673}{2020}\)

Answer 5

\(A=\frac{1}{3}\times\left(\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{2017\times2020}\right)\)

\(\Rightarrow A=\frac{1}{3}\times\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2020}\right)\)

\(\Rightarrow A=\frac{1}{3}\times\left(1-\frac{1}{2020}\right)\)

\(\Rightarrow A=\frac{1}{3}\times\frac{2019}{2020}=\frac{673}{2020}\)