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The question asks one to find an equation to describe the change in surface area of stacks of 1 in by 1 in cubes. The first figure is a single cube, the second figure has a base of two cubes with one cube stacked on the left cube, the second has a base of three cubes with the previous figure stacked on top of it (still keeping the right triangle appearance).

My trouble is in finding a way to describe the change in area through an equation.

Thalia
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1 Answers1

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You're asked to find out an expression for the surface area, why bother with the volume? Anyway, finding the progression of the volume is a trivial problem (just the triangular number sequence).

The surface area is a little harder.

Sketch the diagrams for $n=1,2,3,4,5$. You'll need at least those many cases to be very clear about what's happening. Some cubes have 5 faces "exposed, some have 4, some 3 and some 2. Label them as "5-faces", "4-faces", etc.

Count how many there are in each configuration for these lower values of $n$.

You'll find that for $n \geq 4$,

The number of 5-faces = 2 (constant, since they're the "hanging corner" cubes)

Number of 4-faces = $(n-2) + 1 = n-1$ (a diagonal line at the "exposed surface" plus the single "bottom corner" piece).

Number of 3-faces = $2(n-2)$ (the left and right "edge rows")

Number of 2-faces = $T_{n-3}$, where $T_n$ is the $n$-th triangular number. This is a little tricky to see, but you'll see the 2-faces always stacked in a triangular profile with base $n-3$. So the number of 2-faces is $\frac{1}{2}(n-3)(n-2)$.

Now to find the total area, add all those up, weighted by the areas of each type of face. That is,

$$A = (5)(2) + (4)(n-1)+ (3)(2(n-2)) + 2(\frac{1}{2})(n-3)(n-2) \\ \therefore A= n(n+5)$$

Even though you established this expression for $n \geq 4$, you can quickly verify that it also holds for $n=1,2,3$. Therefore, you have a truly general solution.

You can prove it rigorously by induction by considering the change in the number of each type of face from $n \to n+1$. The reasoning would largely be intuitive, based on visualising the profile of the shape, just spelled out a little more formally.

Deepak
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  • Thank you so much! This was very helpful. The only part that is confusing to me is the part about the cubes with 2 exposed faces. Would you mind explaining that again? – Thalia Oct 01 '14 at 01:35
  • Sketch for $n=4$. You'll see there's just one cube with two exposed faces. It's "touching" 4 other cubes, so only 2 faces (toward you and facing into the paper) are exposed. Now sketch for $n=5$. You'll see a triangular "nest" of 2-faces forming, with three of them now. If you try $n=6$, you'll see a larger triangular nest. So the number of 2-faces goes up in a triangular number sequence that's offset by $3$ from the number of base cubes of the overall shape. Clear enough? – Deepak Oct 01 '14 at 02:03
  • Yes, thank you. Just one more question, can you explain where you got (n-2) from for the equations? – Thalia Oct 01 '14 at 03:25
  • For the 3-faces: just count the number of "edge" (excluding "corner" pieces along both sides. You'll find there are $(n-2)$ on each side. So the total is just double that. For the 2-faces: the general formula for the $n$-th triangular number is $\frac{1}{2}(n)(n+1)$. Use an offset of $3$ in this formula to get what I got. – Deepak Oct 01 '14 at 04:00