TITLE: Prove that in a sequence of numbers $49 , 4489 , 444889 , 44448889\ldots$
QUESTION [9 upvotes]: 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.

REPLY [2 votes]: $$ (\frac{2\cdot10^k+1}{3})^2 = \frac{4\cdot10^{2k}+4\cdot10^k+1}{9} = \frac{4\cdot10^{2k}-4+4\cdot10^k-4+9}{9} = $$
$$ 4\cdot\frac{10^{2k}-1}{9}+4\cdot\frac{10^{k}-1}{9}+1$$