$解：(1) 上述變化規(guī)律可表示為{({\sqrt {n}})}^{2}+1=n+1,{S}_{n}={\frac {\sqrt {n}} {2}}$
$(2) 由題意可知，{OA}^{2}_{1}=1，{OA}^{2}_{2}={({\sqrt {1}})}^{2}+1=2，$
${OA}^{2}_{3}={({\sqrt {2}})}^{2}+1=3，···，{OA}^{2}_{n}={({\sqrt {n-1}})}^{2}+1=n$
$∴ {OA}^{2}_{10}=10$
$∵ {OA}_{10}＞0$
$∴ {OA}_{10}={\sqrt {10}}$
$(3) 由題意可知，{S}^{2}_{1}={\frac {1} {4}}，{S}^{2}_{2}={\frac {2} {4}}，{S}^{2}_{3}={\frac {3} {4}}，···，{S}^{2}_{n}={\frac {n} {4}}$
$∴ {S}^{2}_{1}+{S}^{2}_{2}+{S}^{2}_{3}+···+{S}^{2}_{10}={\frac {1+2+3+···+10} {4}}={\frac {55} {4}}$
