$解：(1)①S_2-S_1=(1+2\sqrt {2})²-(1+\sqrt {2})²=6+2\sqrt {2}$
$②S_3-S_2=(1+3\sqrt {2})²-(1+2\sqrt {2})²=10+2\sqrt {2}$
$③S_4-S_3=(1+4\sqrt {2})²-(1+3\sqrt {2})²=14+2\sqrt {2}$
$(2)S_{n+1}-S_{n}=2(2n+1)+2\sqrt {2}$
$S_{n+1}-S_{n}=[1+(n+1)\sqrt {2}]²-(1+\sqrt {2}n)²$
$=1+2\sqrt {2}(n+1)+2(n+1)-1-2\sqrt {2}n-2n²$
$=2(2n+1)+2\sqrt {2}$
