1.阅读下面的材料,并解答后面的问题:
$\frac{1}{\sqrt{2}+1}$=$\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}$=$\sqrt{2}$-1
$\frac{1}{\sqrt{3}+\sqrt{2}}$=$\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$=$\sqrt{3}$-$\sqrt{2}$;
$\frac{1}{\sqrt{4}+\sqrt{3}}$=$\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{4}+\sqrt{3})(\sqrt{4}-\sqrt{3})}$=$\sqrt{4}$-$\sqrt{3}$
(1)观察上面的等式,请直接写出$\frac{1}{\sqrt{n+1}+\sqrt{n}}$(n为正整数)的结果$\sqrt{n+1}$-$\sqrt{n}$;
(2)计算($\sqrt{n+1}+\sqrt{n}$)($\sqrt{n+1}-\sqrt{n}$)=1;
(3)请利用上面的规律及解法计算:($\frac{1}{\sqrt{2}+1}$+$\frac{1}{\sqrt{3}+\sqrt{2}}$+$\frac{1}{\sqrt{4}+\sqrt{3}}$+…+$\frac{1}{\sqrt{2017}+\sqrt{2016}}$)($\sqrt{2017}+1$).