Find the volume at \(i=\infty\)		Find the surface area at \(i=\infty\)
$$\frac{4\pi \cdot R^{3}}{3} + \sum_{n=1}^{i} \frac{24\pi \cdot 5^{n-1} \cdot R^{3}}{3\cdot2^{3n}}$$	:Initial Rule	$$4\pi \cdot R^{2} + \sum_{n=1}^{i} 24\pi \cdot 5^{n-1} \cdot \frac{R^{2}}{2^{2n}}$$	:Initial Rule
$$\frac{4\pi \cdot R^{3}}{3} + \frac{24\pi \cdot R^{3}}{3} \cdot \sum_{n=1}^{i} \frac{5^{n-1}}{2^{3n}}$$	$$\sum_{k=1}^{n}C \cdot k = a \cdot \sum_{k=1}^{n}k $$	$$4\pi \cdot R^{2} + 24\pi \cdot R^{2} \cdot \sum_{n=1}^{i} \frac{5^{n-1}}{2^{2n}}$$	$$\sum_{k=1}^{n}C \cdot k = a \cdot \sum_{k=1}^{n}k $$
$$\frac{4\pi \cdot R^{3}}{3} + \frac{24\pi \cdot R^{3}}{3} \cdot \sum_{n=1}^{i} \frac{5^{n-1}}{8^{n}}$$	$$(a^{b})^{c} = a^{b \cdot c}$$	$$4\pi \cdot R^{2} + 24\pi \cdot R^{2} \cdot \sum_{n=1}^{i} \frac{5^{n-1}}{4^{n}}$$	$$(a^{b})^{c} = a^{b \cdot c}$$
$$\frac{4\pi \cdot R^{3}}{3} + \pi \cdot R^{3} \cdot \sum_{n=1}^{i} (\frac{5}{8})^{n-1}$$	$$\sum_{k=1}^{n}C \cdot k = a \cdot \sum_{k=1}^{n}k $$	$$4\pi \cdot R^{2} + 6\pi \cdot R^{2} \cdot \sum_{n=1}^{i} (\frac{5}{4})^{n-1}$$	$$\sum_{k=1}^{n}C \cdot k = a \cdot \sum_{k=1}^{n}k $$
$$\frac{4\pi \cdot R^{3}}{3} + \pi \cdot R^{3} \cdot \lim_{i \to \infty}\sum_{n=1}^{i} (\frac{5}{8})^{n-1} = \frac{4\pi \cdot R^{3}}{3} + \pi \cdot R^{3} \cdot \lim_{i \to \infty}\frac{(\frac{5}{8})^{i}-1}{\frac{5}{8}-1}$$	: Geometric Series	$$4\pi \cdot R^{2} + 6\pi \cdot R^{2} \cdot \lim_{i \to \infty}\sum_{n=1}^{i} (\frac{5}{4})^{n-1} = 4\pi \cdot R^{2} + 6\pi \cdot R^{2} \cdot \lim_{i \to \infty}\frac{(\frac{5}{4})^{i}-1}{\frac{5}{4}-1}$$	: Geometric Series
$$\frac{4\pi \cdot R^{3}}{3} + \pi \cdot R^{3} \cdot \frac{0-1}{\frac{5}{8}-1}$$	: Simplify	$$4\pi \cdot R^{2} + 6\pi \cdot R^{2} \cdot \frac{\infty-1}{\frac{5}{4}-1}$$	: Simplify
$$\frac{4\pi \cdot R^{3}}{3} + \pi \cdot R^{3} \cdot \frac{8}{3}$$	: Calculations QED	$$\infty$$	: Calculations QED