Cantor Problem-- Finding the sum at $n=20$

Question

This problem is driving me absolutely nuts.

You have a line segment equal to $1$. You take out the middle third. Can we create a formula for this?

Answer 1

We can do this for the first few $n\geq 0$ then notice a pattern. Let $\ell (n)$ denote the length of line segments at step $n$. Then $$[0,1] \quad , \qquad\ell(0) = 1 $$ $$[0,\frac{1}{3}] \cup [\frac{2}{3},1] \quad , \qquad\ell(1) = 2(\frac{1}{3})$$ $$\left([0,\frac{1}{9}]\cup [\frac{2}{9},\frac{1}{3}] \right)\cup \left([\frac{2}{3},\frac{7}{9}]\cup [\frac{8}{9},1]\right) \quad , \qquad\ell(2) = 4(\frac{1}{3^2})$$ $$\vdots$$

You can try the next iteration, and find that

At every step, the number of line segments doubles, and the length of each segment is $3$-times smaller. Hence at each step $n$ starting with $\mathbf{n=0}$, the length of all line segments at that step is given by $$ \ell(n) = 2^n \left( \frac{1}{3} \right)^n$$

Just plug in $n$ to get the answer (note that if the question assumes the starting position to be $n=0$, you will need to find $\ell(20)$. If the question assumes the starting position to be $n=1$, you will need to find $\ell(19)$)