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Question Number 152946 by john_santu last updated on 03/Sep/21

$$\frac{4038}{1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{1+2+3+4+...+2019}}=?$$

Commented by mathdanisur last updated on 03/Sep/21

$$▴\frac{2\cdot 2019}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2019}}$$$$\mathbf{A}=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2019}$$$$\mathbf{A}=1+\frac{2}{2\cdot 3}+\frac{2}{3\cdot 4}+...+\frac{2}{2019\cdot 2020}$$$$\frac{\mathbf{A}}{2}=\frac{1}{2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{2019\cdot 2020}$$$$\frac{\mathbf{A}}{2}=\frac{1}{2}+\frac{3-2}{2\cdot 3}+\frac{4-3}{3\cdot 4}+...+\frac{2020-2019}{2019\cdot 2020}$$$$\frac{\mathbf{A}}{2}=\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}$$$$\frac{\mathbf{A}}{2}=1-\frac{1}{2020}\Rightarrow \mathbf{A}=\frac{2\cdot 2019}{2020}$$$$\Rightarrow \frac{\frac{2\cdot 2019}{2\cdot 2019}}{2020}=2020$$

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