Sum of square of Roots

Firstly, we note that $\sqrt{a^2 + a\sqrt{4a^2 + a\sqrt{16a^2 + a\sqrt{64a^2 + a\sqrt{...}}}}} \; = \; \kappa a$ where $\kappa \; = \; \sqrt{1 + \sqrt{4 + \sqrt{16 + \sqrt{64 + \sqrt{...}}}}} \approx 1.78$ Now $\begin{aligned} \sum_{a_=1}^{a_2} a_1& = \; \tfrac12a_2(a_2+1) \\ \sum_{a_2=1}^{a_3}\tfrac12a_2(a_2+1) & = \; \tfrac16a_3(a_3+1)(a_3+2) \\ \sum_{a_3=1}^{a_4}\tfrac16a_3(a_3+1)(a_3+2) & = \; \tfrac{1}{4!}a_4(a_4+1)(a_4+2)(a_4+3) \end{aligned}$ Continuing this pattern inductively, we obtain $\begin{aligned} \sum_{a_{99}=1}^{a_{100}} \sum_{a_{98}=1}^{a_{99}} \cdots \sum_{a_2=1}^{a_3} \sum_{a_1=1}^{a_2} a_1 & = \; \tfrac{1}{100!} a_{100}(a_{100}+1)(a_{100}+2)\cdots (a_{100}+99) \\ f(x) & = \; \tfrac{\kappa}{101!} x(x+1)(x+2)(x+3)\cdots (x+99)(x+100) \end{aligned}$ This is defined for positive integer $x$ only, but we can consider the function $f$ defined by this formula for any real $x$ . Then the zeros of $f(x)$ are $0,-1,-2,...,-100$ , so the desired answer is $0^2 + (-1)^2 + (-2)^2 + \cdots + (-100)^2 \; = \; \boxed{338350}.$