Fractal Cubes

Relevant wiki: General Term Pattern Recognition

For every new cube formed, there will be 4 new square areas arising from each face (Imagine a rising square platform, revealing new 4 lateral squares), and from $a_{3}$ on, there will be 5 faces on each old cube for such formation can occur. Hence, the recursion can be evaluated as followed:

$a_{1} = 6\times (3\times 3) = 54$

$a_{2} = 54 + 6\times 4\times (1^2) = 78$

$a_{3} = 78 + 6\times 5\times 4\times (\dfrac{1}{3})^2$

$a_{4} = 78 + 6\times 5\times 4\times (\dfrac{1}{3})^2 + 6\times 5^2 \times 4\times (\dfrac{1}{9})^2$

$\vdots$

$a_{n} = 78 + 6\times 4[ (\dfrac{5}{9}) + (\dfrac{5}{9})^2 + \cdots + (\dfrac{5}{9})^{n-2}]$

Let $S = (\dfrac{5}{9}) + (\dfrac{5}{9})^2 + \cdots + (\dfrac{5}{9})^{n-2} = \dfrac{\dfrac{5}{9}}{1-\dfrac{5}{9}} = \dfrac{5}{4}$ . (This is a sum of a geometric progression)

Hence, $\displaystyle \lim_{n \to \infty} a_{n} = 78 + 6\times 4(\dfrac{5}{4}) = \boxed{108}$ .