$1/4$ is in the Cantor set?

$1/4$ is in the Cantor set. It is in the lower third. And it is in the upper third of the lower third. And in the lower third of that, and in the upper third of that, and so on.

The quickest way to see this is that it is exactly $1/4$ of the way from $1/3$ down to $0$, and then use self-similarity and symmetry.

A more plodding way to show it is to look at the series $$ \frac29 + \frac{2}{9^2}+\frac{2}{9^3}+\cdots=\frac14. $$ This shows that the base-$3$ expansion of $1/4$ is $0.02020202\ldots$. Since it has a base-$3$ expansion with only $0$s and $2$s, it is in the Cantor set.

Hint: write $\frac{1}{4}$ in base $3$.

To do this, just observe that

$$\frac{1}{4}=\frac{2}{9-1}=\frac{2}{9}\frac{1}{1-\frac{1}{9}}$$

and write the geometric series which has that limit.

We use mathematical induction to show $\frac{1}{4}$ and $\frac{3}{4}$ are in $C_n,\, \forall n\in \mathbb{N}\cup \{0\}$.

Clearly, they are in $[0,1]$.

Suppose they are in $C_N$. We observe the structure of $[0,\frac{1}{3}]$ after $n+1$ cuts is similar to $[0,1]$ after $n$ cuts. Hence if $x \in C_N$, then $\frac{x}{3} \in C_{N+1}$. Since by assumption, $\frac{3}{4} \in C_{N}$, we have $\frac{1}{4} \in C_{N+1}$. Moreover, since the cut is symmetry on $[0,1]$, we know $\frac{3}{4} \in C_{N+1}$.

Hence by induction, $\frac{1}{4}$ and $\frac{3}{4}$ are in $C_n,\, \forall n\in \mathbb{N}$, we have $\frac{1}{4}$ and $\frac{3}{4}$ are in $C$.

From the construction of the Cantor set using "deleted thirds", one sees that each of the numbers $$\textstyle a_1={1\over3},\ \ a_2={1\over3}-{1\over9},\ \ a_3={1\over3}-{1\over9}+{1\over27}, \ \ \ldots$$ is an element of the Cantor set.

Since the sequence $(a_k)$ is the sequence of partial sums of the convergent Geometric series $\sum\limits_{n=1}^\infty(-1)^{n+1} ({1\over3^n}) $, it follows that
$$\lim_{k\rightarrow\infty} a_k= \sum_{n=1}^\infty(-1)^{n+1} ({\textstyle{1\over3^n})}= -\sum_{n=1}^\infty (\textstyle{-1\over3})^n=(-1)\cdot{-1/3\over 1-(-1/3)} ={1/4}.$$

Now, the Cantor set is closed; and, since the Cantor set contains each $a_k$, it must contain the limit of $(a_k)$. Thus, the Cantor set contains the point $1/4$.

(This argument shouldn't seem so mysterious if you determine where $1/4$ lies in relation to the $a_k$; see the first paragraph of Michael Hardy's answer.)