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观察图3，并认真分析下列各式，然后解答问题.
$(\sqrt{1})^2 +1=2, S_1=\frac{\sqrt{1}}{2};$
$(\sqrt{2})^2 +1=3, S_2=\frac{\sqrt{2}}{2};$
$(\sqrt{3})^2 +1=4, S_3=\frac{\sqrt{3}}{2};$
……
(1)用含有$n$($n$是正整数)的等式表示上述变化规律；
(2)求$OA_{10}$的长；
(3)求$S_1^2 + S_2^2 + S_3^2 + \dots + S_{10}^2$的值.
$(\sqrt{1})^2 +1=2, S_1=\frac{\sqrt{1}}{2};$
$(\sqrt{2})^2 +1=3, S_2=\frac{\sqrt{2}}{2};$
$(\sqrt{3})^2 +1=4, S_3=\frac{\sqrt{3}}{2};$
……
(1)用含有$n$($n$是正整数)的等式表示上述变化规律；
(2)求$OA_{10}$的长；
(3)求$S_1^2 + S_2^2 + S_3^2 + \dots + S_{10}^2$的值.
答案
(1) $(\sqrt{n})^2+1=n+1,S_n=\dfrac{\sqrt{n}}{2}$.
(2) $\sqrt{10}$
(3) $S_1^2+S_2^2+S_3^2+\dots+S_{10}^2=(\dfrac{\sqrt{1}}{2})^2+(\dfrac{\sqrt{2}}{2})^2+(\dfrac{\sqrt{3}}{2})^2+\dots+(\dfrac{\sqrt{10}}{2})^2=\dfrac{1}{4}×(1+2+3+\dots+10)=\dfrac{55}{4}$.
(2) $\sqrt{10}$
(3) $S_1^2+S_2^2+S_3^2+\dots+S_{10}^2=(\dfrac{\sqrt{1}}{2})^2+(\dfrac{\sqrt{2}}{2})^2+(\dfrac{\sqrt{3}}{2})^2+\dots+(\dfrac{\sqrt{10}}{2})^2=\dfrac{1}{4}×(1+2+3+\dots+10)=\dfrac{55}{4}$.
