Nested Square Pyramids and Spheres!

$V_{s}(1) = \dfrac{4}{3}\pi R_{1}^3$ and $V_{p}(1) = \dfrac{1}{3}x_{2}^2H_{1}$

Using the above diagram we have:

$(H_{1} - R_{1})^2 + \dfrac{x_{1}^2}{2} = R_{1}^2 \implies x_{1}^2 = 2(2H_{1}R_{1} - H_{1}^2) \implies$ $V_{p}(1) = \dfrac{2}{3}(2H_{1}^2R_{1} - H_{1}^3) \implies$

$\dfrac{dV_{p}(1)}{dH_{1}} = \dfrac{2}{3}H_{1}(4R_{1} - 3H_{1}) = 0$ and $H_{1} \neq 0 \implies H_{1} = \dfrac{4}{3}R_{1} \implies x_{1} = \dfrac{4}{3}R_{1} = H_{1}$ .

Let $S = S_{1}$ .

$S_{1} = \dfrac{1}{2}\sqrt{x_{1}^2 + 4H_{1}^2} = \dfrac{2}{3}\sqrt{5}R_{1}$

and the area of the blue triangle $A = \dfrac{1}{2}x_{1}H_{1} = \dfrac{x_{1} + 2S_{1}}{2}R_{2} \implies$ $\boxed{R_{2} = \dfrac{x_{1}H_{1}}{x_{1} + 2S_{1}} = \dfrac{4}{3}(\dfrac{1}{1 + \sqrt{5}})R_{1} = \dfrac{2}{3\phi}R_{1}}$ , where $\phi = \dfrac{1 + \sqrt{5}}{2}$ .

$H_{2} = \dfrac{4}{3}R_{2} = \dfrac{8}{9\phi}R_{1} = x_{2}$ and $S_{2} = \dfrac{1}{2}\sqrt{x_{2} + 4H_{2}} = \dfrac{4}{9\phi}\sqrt{5}R_{1} \implies$

$\boxed{R_{3} = \dfrac{x_{2}H_{2}}{x_{2} + 2S_{2}} = \dfrac{8}{9\phi}\dfrac{1}{(1 + \sqrt{5})}R_{1} = \dfrac{4}{9\phi^2}R_{1} = (\dfrac{2}{3\phi})^2R_{1}}$

In General:

$\boxed{R_{n} = (\dfrac{2}{3\phi})^{n - 1}R_{1}}$

Note: $x_{n} = H_{n} = \dfrac{4}{3}(\dfrac{2}{3\phi})^{n - 1}R_{1}$ and $S_{n} = \dfrac{2}{3}(\dfrac{2}{3\phi})^{n - 1}\sqrt{5}R_{1}$

$\implies V_{s}(n) = \dfrac{4}{3}\pi R_{n}^3 = (\dfrac{8}{27\phi^3})^{n - 1} * V_{s}(1)$ $\implies T = \sum_{n = 1}^{\infty} V_{s}(n) =$

$V_{s}(1)\sum_{n = 1}^{\infty} (\dfrac{8}{27\phi})^{n - 1} =$ $\dfrac{27\phi^3}{27\phi^3 - 8}V_{s}(1) = \dfrac{(3\phi)^3}{(3\phi)^3 - 2^3}V_{s}(1) \implies$

$\dfrac{T}{V_{s}(1)} = \dfrac{(3\phi)^3}{(3\phi)^3 - 2^3} = \dfrac{(a\phi)^a}{(a\phi)^a - b^a}$ $\implies a + b = \boxed{5}$ .