<s>[INST] 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. 
 [/INST] Without words:
$$\begin{align}
\left(6\frac{10^k-1}{9}+1\right)^2 &= 36 \frac{10^{2k} - 2\cdot 10^k + 1}{81} + 12\frac{10^k-1}{9} + 1\\
&= 4\frac{10^k-1}{9}\cdot 10^k - 4 \frac{10^k-1}{9} + 12 \frac{10^k-1}{9} + 1\\
&=  4\frac{10^k-1}{9}\cdot 10^k + 8 \frac{10^k-1}{9} + 1.
\end{align}$$
</s>