Question

Tìm số tự nhiên n biết rằng \(\frac{1}{\sqrt{1^3+2^3}}+\frac{1}{\sqrt{1^3+2^3+3^3}}+...+\frac{1}{\sqrt{1^3+2^3+...+n^3}}=\frac{2017}{2019}\)

Answer 1

\(1^3+2^3+...+n^3=\left(1+2+...+n\right)^2\) (có thể chứng minh bằng quy nạp)

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

\(\Rightarrow\frac{1}{\sqrt{1^3+2^3+...+n^3}}=\frac{1}{\sqrt{\left(\frac{n\left(n+1\right)}{2}\right)^2}}=\frac{2}{n\left(n+1\right)}=2\left(\frac{1}{n}-\frac{1}{n+1}\right)\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\frac{2}{n+1}=1-\frac{2017}{2019}=\frac{2}{2019}\)

\(\Rightarrow n+1=2019\Rightarrow n=2018\)