2*2019/((1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+2019))). Created by fk040826. matematika-ru

Ответ:2020Пошаговое объяснение: Тогда:

(1+2+3+...+2019)))...

fk040826 @fk040826

April 2022 1 79 Report

2*2019/((1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+2019)))

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igorShap

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Ответ:

2020

Пошаговое объяснение:

$\dfrac{1}{1+2+...+k}=\dfrac{1}{\frac{1+k}{2}k}=\dfrac{2}{k(k+1)}$

Тогда: $\dfrac{2*2019}{1+\frac{1}{1+2}+...+\frac{1}{1+...+2019}}=\dfrac{2*2019}{\frac{2}{1*2}+\frac{2}{2*3}+...+\frac{2}{2019*2020}}=\dfrac{2019}{\frac{1}{1*2}+\frac{1}{2*3}+...+\frac{1}{2019*2020}}=\dfrac{2019}{\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2020}}=\dfrac{2019}{1-\frac{1}{2020}}=\dfrac{2019}{\frac{2019}{2020}}=2020$

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