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. 

$44...488...89$ 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.
