It is well known that a line has 2 points.
A square has 4 lines.
A cube has 6 squares.
A tesseract has 8 cubes.
And continuing there will be 10, 12, 14, etc.
Why does the sequence increase by 2? I can figure out why the number of points in each case is always $2^N$ (because you are doubling the points in each instance), but can't figure out why in each dimension N the number of (N-1) dimension figures used is 2n.
Method 1: See http://en.wikipedia.org/wiki/Hypercube. This says that
Why?
Consider the $n$-dimension hypercube with vertices at $(\pm x,\pm x,\ldots,\pm x)$. Its volume $V_n(x)$ is $(2x)^n$. Expand the hypercube a little bit from $x$ to $x + \epsilon$, each $m$-dimension face of the hypercube will contribute an extra hypervolume $V_m(x) \epsilon^{n-m}$ to the new hypercube. This means:
$$V_{n}(x+\epsilon) = \sum_{m=0}^n N_{n,m} V_{m}(x)\epsilon^{n-m}$$ where $N_{n,m}$ is the number of $m$-dimension "faces" of a $n$-dimension hypercube.
One the other hand, binomial theorem tell us:
$$V_n(x+\epsilon) = 2^n(x+\epsilon)^n = 2^n\sum_{m=0}^n \binom{n}{m} x^{m} \epsilon^{n-m} =\sum_{m=0}^n 2^{n-m}\epsilon{n}{m}V_{m}(x)\epsilon^{n-m} $$
By comparing the coefficients of powers in $\epsilon$, we immediately get:
$$\boxed{N_{n,m} = 2^{n-m} \binom{n}{m}}$$
Method 2: It follows from the Euler characteristic (http://en.wikipedia.org/wiki/Euler_characteristic).
Let $b_n$ be the $n$-th Betti number, i.e., the rank of the $n$-th singular homology group. Then, the Euler characteristic is $$\chi=\sum^\infty_{n=0}(-1)^nb_n$$ Thus, use Euler's formula: $$F=\sum^\infty_{n=0}(-1)^nb_n+\text{number of edges $-$ number of vertices}$$