User blog comment:Edwin Shade/A Mengol/@comment-1605058-20171026043224/@comment-1605058-20171026195318

Wikipedia lists the formula for the surface area at each iteration. However, there are multiple ways to find out the closed form formula just from the recurrence. Let me mention two of them.

If you know nothing about recurrences, you can think this way: since \(S_n\) involves \(20\cdot S_{n-1}\), intuitively \(S_n\) should grow roughly like some constant times \(20^n\) (the \(8^n\) term will negligible). To find this constant, we can write \(S_n=20^n\cdot T_n\), then, plugging this into the recurrence and dividing, we find \(T_n=T_{n-1}-6\cdot\left(\frac{8}{20}\right)^n\), and iterating this formula we get \(T_n=T_0-6\cdot\left(\frac{8}{20}\right)^1-6\cdot\left(\frac{8}{20}\right)^2-\dots-6\cdot\left(\frac{8}{20}\right)^n\). \(T_0=6\) and we can use the formula for the sum of geometric series, this finally gives \(T_n=6+4\left(\left(\frac{8}{20}\right)^n-1\right)\), which finally gives, multiplying by \(20^n\), \(S_n=2\cdot 20^n+4\cdot 8^n\).

The other method is slightly more complicated, but much more general and powerful, and it's the method of generating functions. Namely, we consider a power series \(F(x)=\sum_{n=0}^\infty S_nx^n\), where \(x\) is a (formal) variable. If we plug in the recurrence formula, we find \(F(x)=S_0+\sum_{n=1}^\infty(20S_{n-1}-6\cdot 8^n)x^n=6+\sum_{n=1}^\infty S_{n-1}x^n+\sum_{n=1}^\infty(8x)^n=6+20xF(x)-6\cdot\frac{8x}{1-8x}\). We rearrange this to get \(F(x)=\frac{6}{1-20x}-\frac{48x}{(1-8x)(1-20x)}=\frac{2}{1-20x}+\frac{4}{1-8x}\) (the last step is the so called partial fraction decomposition). This last sum can be written as a power series too: \(2\sum_{n=0}^\infty 20^nx^n+4\sum_{n=0}^\infty 8^nx^n=2\sum_{n=0}^\infty (2\cdot 20^n+4\cdot 8^n)x^n\). Comparing the coefficients of the two power series gives \(S_n=2\cdot 20^n+4\cdot 8^n\).

or, you know, you can ask Wolfram Alpha...