Nice series!

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$\sqrt{a_{2n-1}+a_{2n+1}}=\sqrt{2+8(\dfrac{1}{(2n-1)^{2}}+\dfrac{1}{(2n+1)^{2}})}$

$=\sqrt{2} \sqrt{1+8\dfrac{4n^{2}+1}{(4n^{2}-1)^{2}}}$ let $4n^{2}=a$

Then,

$\sqrt{a_{2n-1}+a_{2n+1}}=\sqrt{2} \sqrt{1+8\dfrac{a+1}{(a-1)^{2}}}$

$=\sqrt{2} \sqrt{\dfrac{a^{2}-2a+1+8a+8}{(a-1)^{2}}}=\sqrt{2} \times \dfrac{a+3}{a-1}$

$=\sqrt{2} \times (1+\dfrac{4}{a-1})$ Now, substitute a back to the equation.

$=\sqrt{2} \times (1+2(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}))$

Since the summation of the fraction term is a telescoping sum , so, the sum is

$\Sigma_{n=1}^{520} 2(\dfrac{1}{2n-1}-\dfrac{1}{2n+1})=2(1-\dfrac{1}{1041})$

So, the whole summation is

$Sum=\sqrt{2}(520+2(1-\dfrac{1}{1041}))=\boxed{738.22}$