Question

Tìm số tự nhiên n , biết rằng : \(\dfrac{1}{\sqrt{1^3+2^3}}+\dfrac{1}{\sqrt{1^3+2^3+3^3}}+...+\dfrac{1}{\sqrt{1^3+2^3+3^3+...+n^3}}=\dfrac{2015}{2017}\)

Answer 1

\(\sqrt{1^3+2^3}=\sqrt{\left(1+2\right)^2}=3\)

\(\sqrt{1^3+2^3+3^3}=\sqrt{\left(1+2+3\right)^2}=1+2+3=6\)

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

=>\(\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{1+2+...+n}=\dfrac{2015}{2017}\)

\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{2015}{2017}\)

\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{2015}{2017}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{2015}{4034}\)

=>1/(n+1)=1/2017

=>n+1=2017

=>n=2016