Ta có:[tex]\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{(n+1)^2n-n^2(n+1)}[/tex]
[tex]=\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{(n+1)^2n-n^2(n+1)}=\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)(n+1-n)}[/tex]
[tex]=\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)}=\frac{\sqrt{n}}{n}-\frac{\sqrt{n+1}}{n+1}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}[/tex]
Áp dụng đẳng thức trên ta có:[tex]\frac{1}{2.\sqrt{1}+1.\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+.....+\frac{1}{2017\sqrt{2016}+2016\sqrt{2017}}[/tex]
=[tex]1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+......+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}[/tex]
[tex]=1-\frac{1}{\sqrt{2017}}[/tex]