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. 
 A: $44...488...89$ 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.
A: Without words:
$$\begin{align}
\left(6\frac{10^k-1}{9}+1\right)^2 &= 36 \frac{10^{2k} - 2\cdot 10^k + 1}{81} + 12\frac{10^k-1}{9} + 1\\
&= 4\frac{10^k-1}{9}\cdot 10^k - 4 \frac{10^k-1}{9} + 12 \frac{10^k-1}{9} + 1\\
&=  4\frac{10^k-1}{9}\cdot 10^k + 8 \frac{10^k-1}{9} + 1.
\end{align}$$
A: $$ (\frac{2\cdot10^k+1}{3})^2 = \frac{4\cdot10^{2k}+4\cdot10^k+1}{9} = \frac{4\cdot10^{2k}-4+4\cdot10^k-4+9}{9} = $$
$$ 4\cdot\frac{10^{2k}-1}{9}+4\cdot\frac{10^{k}-1}{9}+1$$
A: Let $m=111111\dots$ the decimal number consisting of $n$ consecutive '$1$'s so that $9m+1=10^n$
Then the given number is $$10^nm+8m+1=(9m+1)m+8m+1=36m^2+12m+1=(6m+1)^2$$
