$ 解:(1)\sqrt {1+\frac {1}{4^{2}}+\frac {1}{5^{2}}}=1+\frac {1}{4}-\frac {1}{4+1}=1\frac {1}{20}$
$\sqrt {1+\frac {1}{4^{2}}+\frac {1}{5^{2}}}=\sqrt {\frac {441}{400}}=\frac {21}{20}=1\frac {1}{20}$
$(2)\sqrt {1+\frac {1}{n^{2}}+\frac {1}{(n+1)^{2}}}=1+\frac {1}{n}-\frac {1}{n+1}=1+\frac {1}{n(n+1)}$
