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Question Number 153839 by liberty last updated on 11/Sep/21

Commented by mathdanisur last updated on 11/Sep/21

$$\mathrm{x}(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+4041})=4041$$$$▴\mathbf{A}=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+\mathbf{n}}=\frac{2\mathbf{n}}{\mathbf{n}+1}$$$$\mathbf{n}=4041\Rightarrow \mathbf{A}=\frac{2\cdot 4041}{4041+1}=\frac{2\cdot 4041}{2\cdot 2021}=\frac{4041}{2021}$$$$\Rightarrow \mathbf{x}\cdot \frac{4041}{2021}=4041\Rightarrow \mathbf{x}=2021▴$$

Answered by puissant last updated on 11/Sep/21

$$1+\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+4041}$$$$=\frac{1}{\frac{4042}{2\times 4041}}=\frac{2\times 4041}{4042}..$$$$\Rightarrow x(1+\frac{1}{1+2}+....+\frac{1}{1+2+..+4041})=4041$$$$\Rightarrow \frac{\left(2\times 4041\right)x}{4042}=4041$$$$\Rightarrow \frac{2x}{4042}=1\Rightarrow 2x=4042\Rightarrow x=2021..$$$$\therefore \because x=2021....◼$$

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