500

$Let\quad x=\sqrt { 500+\sqrt { 500+\sqrt { 500+... } } } \\ \\ { x }^{ 2 }=500+\sqrt { 500+\sqrt { 500+... } } \\ { x }^{ 2 }=500+x\\ { x }^{ 2 }-x-500=0\\ { (x-\frac { 1 }{ 2 } ) }^{ 2 }-\frac { 1 }{ 4 } -\frac { 2000 }{ 4 } =0\\ x=\frac { 1\pm \sqrt { 2001 } }{ 2 } \\ x=\frac { 1+\sqrt { 2001 } }{ 2 } \quad (x>0)\\ \\ \frac { 1+\sqrt { 2001 } }{ 2 } \quad is\quad irreducible\quad and\quad is\quad in\quad the\quad form\\ \frac { a+\sqrt { b } }{ c } .\\ \\ a=1,\quad b=2001,\quad c=2\\ \\ \therefore \boxed { a+b-c=1+2001-2=2000 }$

$x=\sqrt{n+\sqrt{n+\sqrt{n+...}}}$ $x=\sqrt{n+x}$ Here, you get the quadratic equation: $x^{2}-x-n$ By the quadratic formula, we generate the equation $x=\dfrac{1+\sqrt{1+4n}}{2}$ Substitue $n=500$ to get $\implies \dfrac{1+\sqrt{2001}}{2} \implies 1+2001-2=\boxed{2000}$