Question

\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}\)+\(\dfrac{1}{3\sqrt{2}+2\sqrt{3}}\)+....+\(\dfrac{1}{400\sqrt{399}+399\sqrt{400}}\)

Answer 1

Ta có công thức tổng quát là \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(n+1-n\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

Vậy \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{400\sqrt{399}+399\sqrt{400}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{399}}-\dfrac{1}{\sqrt{400}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{400}}=1-\dfrac{1}{20}=\dfrac{19}{20}\)