# 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.

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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.