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Question Number 108365 by mathdave last updated on 16/Aug/20

Answered by JDamian last updated on 16/Aug/20

$$S=\underset{n=1}{\sum ^{2019}}\frac{1+2+3+\cdot \cdot \cdot +n}{1^3+2^3+3^3+\cdot \cdot \cdot +{2019}^3}=$$$$=\frac{\underset{n=1}{\sum ^{2019}}(1+2+\cdot \cdot \cdot +n)}{1^3+2^3+3^3+\cdot \cdot \cdot +{2019}^3}$$$$1+2+\cdot \cdot \cdot +n=\frac{n(n+1)}{2}$$$$1^3+2^3+\cdot \cdot \cdot +{2019}^3={\left(\frac{2019\cdot 2020}{2}\right)}^2$$$$S={\left(\frac{2}{2019\cdot 2020}\right)}^2\cdot \frac{1}{2}(\underset{n=1}{\sum ^{2019}}{n}^2+\underset{n=1}{\sum ^{2019}}n)$$$$\underset{n=1}{\sum ^{2019}}n=\frac{2019\cdot 2020}{2}$$$$\underset{n=1}{\sum ^{2019}}{n}^2=\frac{2019\cdot 2020\cdot 4039}{6}$$$$S=\frac{1}{2019\cdot 2020}\cdot (1+\frac{4039}{3})$$$$$$

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