Question

Tính

\(B=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{2\sqrt{3}+3\sqrt{2}}+\dfrac{...1}{99\sqrt{100}+100\sqrt{99}}\)

Answer 1

Ta có công thức tổng quát

\(\dfrac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)Vậy \(P=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{2\sqrt{3}+3\sqrt{2}}+...+\dfrac{1}{99\sqrt{100}+100\sqrt{99}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{100}}=1-\dfrac{1}{10}=\dfrac{9}{10}\)