Question

[galaxyLâm]

Bài 1: Tìm  \(n\inℕ^∗\)

        biết  :     \(2n:\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+.....+\frac{1}{1+2+...+n}\right)=2020\)

Answer 1

[Nguyễn Linh Chi]

Ta có: 

\(2n:\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+.....+\frac{1}{1+2+...+n}\right)=2020\)

<=> \(2n:\left(\frac{2}{2}+\frac{2}{3.2}+\frac{2}{4.3}+...+\frac{2}{\left(n+1\right).n}\right)=2020\)

<=> \(n:\left(1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\right)=2020\)

<=> \(n:\left(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)=2020\)

<=> \(n:\left(1-\frac{1}{n+1}\right)=2020\)

<=> \(n:\frac{n}{n+1}=2020\)

<=> n + 1 = 2020

<=> n = 2019