[tex]\it \dfrac{\sqrt3+\sqrt{12} +\sqrt{27} +\sqrt{48}+\ ...\ +\sqrt{300}}{\sqrt{75}} =
\\ \\ \\
=\dfrac{\sqrt3+2\sqrt3+3\sqrt3+4\sqrt3+\ ...\ +100\sqrt3}{\sqrt{25\cdot3}} =
\\ \\ \\
=\dfrac{\sqrt3(1+2+3+4+\ ...\ +100)}{5\sqrt3} = \dfrac{1+2+3+4+\ ...\ +100}{5}=[/tex]
[tex]\it = \dfrac{\dfrac{100\cdot101}{2}}{5} = \dfrac{50\cdot101}{5} = 10\cdot101 = 1010[/tex]
