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第113页

信息发布者：

解：$ (1) $两个电视塔的传播半径分别为$r_{1} = \sqrt {2Rh_{1}}，$$r_{2} = \sqrt {2Rh_{2}}，$

则半径之比为$\frac {r_{1}}{r_{2}} = \frac {\sqrt {2Rh_{1}}}{\sqrt {2Rh_{2}}} = \sqrt {\frac {h_{1}}{h_{2}}} = \frac {\sqrt {h_{1}\ \mathrm {h}_{2}}}{h_{2}}；$

$ (2) $电视塔高$h = 120m = 0.12\ \mathrm {km}，$

传播半径$r = \sqrt {2 ×6400 ×0.12} = \sqrt {1536} = 16\sqrt {6}\mathrm {km}$

 证明：$\frac{1}{\sqrt{n+1} - \sqrt{n}} = \frac{\sqrt{n+1} + \sqrt{n}}{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})} = \frac{\sqrt{n+1} + \sqrt{n}}{(n+1) - n} = \sqrt{n+1} + \sqrt{n}，$

所以$\sqrt{n+1} - \sqrt{n}$的倒数是$\sqrt{n+1} + \sqrt{n}$