Is this the appropriate way to show that this language is not context-free?
Given the language $L$ containing the words $1$, $101$, $101001$, $1010010001$, where each word $L_n$ is of the form $10^110^21...10^{n-1}10^n1$.
Assume that $L$ is context-free, and as such that there is some $p$ which is the pumping length for $L$.
Consider $s = L_{p-1}$. It is trivial to show that $|s| \geq p$.
By the pumping lemma, we should be able write $s$ in the form $s = > wvxyz$ where $|vxy| \leq p$ and $|vy| \geq 1$ such that $wv^ixy^iz \in > L$ for all $i$.
As $i = 0$, being a possible number of repeats, would give $wxz \in > L$.
Therefore, as $L$ only permits deletion from the tail of words we can derive that $x = \lambda$ and $z = \lambda$.
Also we can see that either $v$ or $y$ must equal $0^{p-1}1$. We can confirm that $|vxy| \leq p$ holds for this.
However, $i = 2$ is also a possible number of repeats, but would result in a word that ends $0^{p-1}10^{p-1}1$ and therefore that is not contained in $L$.
Therefore we have a contradiction, and $L$ is not context-free.
Thankyou for your comments. Yes, this is homework.