Prove that in a sequence of numbers $49 , 4489 , 444889 , 44448889\ldots$

Prove that in a sequence of numbers $49 , 4489 , 444889 , 44448889\ldots$ in which every number is made by inserting $48$ in the middle of previous as indicated, each number is the square of an integer.

Without words:

$$\begin{align} \left(6\frac{10^k-1}{9}+1\right)^2 &= 36 \frac{10^{2k} - 2\cdot 10^k + 1}{81} + 12\frac{10^k-1}{9} + 1\\ &= 4\frac{10^k-1}{9}\cdot 10^k - 4 \frac{10^k-1}{9} + 12 \frac{10^k-1}{9} + 1\\ &= 4\frac{10^k-1}{9}\cdot 10^k + 8 \frac{10^k-1}{9} + 1. \end{align}$$

$44...488...89$ has $n+1$ numbers "$4$", $n$ numbers "$8$", and the "$9$". So:

$$ 44...488...89 = 4\cdot\frac{10^n-1}{9}\cdot 10^{n+1} + 8\cdot\frac{10^n-1}{9}\cdot 10 + 9 $$

Now, we say $10^n = y$ so $$ \begin{align} &\frac{4\cdot (10y-1)\cdot 10y + 8\cdot 10(y-1) + 81}{9}\\ &=\frac{400y^2 + 40y +1}{9}\\ &=\left(\frac{20y + 1}{3}\right)^2 \end{align} $$

Note that as $y = (10^n)$, $3 | (20y +1)$, for any $n$ value.

Let $m=111111\dots$ the decimal number consisting of $n$ consecutive '$1$'s so that $9m+1=10^n$

Then the given number is $$10^nm+8m+1=(9m+1)m+8m+1=36m^2+12m+1=(6m+1)^2$$

$$ (\frac{2\cdot10^k+1}{3})^2 = \frac{4\cdot10^{2k}+4\cdot10^k+1}{9} = \frac{4\cdot10^{2k}-4+4\cdot10^k-4+9}{9} = $$

$$ 4\cdot\frac{10^{2k}-1}{9}+4\cdot\frac{10^{k}-1}{9}+1$$