$\sqrt{a_{n}}=\sqrt{2-\frac{1}{n+\sqrt{n^{2}+\frac{1}{4}}}}$ $=\sqrt{(\sqrt{2n^{2}+2n+1}-\sqrt{2n^{2}-2n+1})^{2}}$ $=\sqrt{2n^{2}+2n+1}-\sqrt{2n^{2}-2n+1}$

For completeness. First, $4n^4+1=4n^4+4n^2+1-4n^2=(2n^2-2n+1)(2n^2+2n+1)=u_n\cdot u_{n+1}$, where $u_n = 2n^2-2n+1$. With this, we have \[ a_n = 2\left(2n^2+1 - \sqrt{4n^4+1}\right) = 2\left(\frac{u_n+u_{n+1}}{2} - \sqrt{u_nu_{n+1}}\right), \]hence \[ \sqrt{a_n} = \sqrt{u_{n+1}}-\sqrt{u_n}. \]So, the given sum telescopes to $\sqrt{u_{120}} - \sqrt{u_1} = 168$.