Prove that 16, 1156, 111556, 11115556, 1111155556… are squares.

Mark Bennet already suggested looking at the numbers as geometric series, so I'll use a slightly different approach. Instead of writing the squares like that, try writing them as follows:

$$\begin{align} 15&.999\ldots = 16 \\ 1155&.999\ldots = 1156 \\ 111555&.999\ldots = 111556 \\ \vdots\end{align}$$

These numbers can be expressed as a sum of three numbers, as follows:

$$\begin{align} 111111&.111\ldots \\ 444&.444\ldots \\ 0&.444\ldots \\ \hline 111555&.999\ldots \end{align}$$

Since $1/9 = 0.111\ldots$, we get

$$\begin{align} 111111&.111\ldots = \frac{1}{9} \cdot 10^{2k} \\ 444&.444\ldots = \frac{1}{9} \cdot 4 \cdot 10^k \\ 0&.444\ldots = \frac{1}{9} \cdot 4 \\ \hline 111555&.999\ldots = \frac{1}{9} \left(10^{2k} + 4 \cdot 10^k + 4\right). \end{align}$$

But this can be written as a square:

$$\frac{1}{9} \left(10^{2k} + 4 \cdot 10^k + 4\right) = \left(\frac{10^k + 2}{3}\right)^2.$$

Since $10^k + 2$ is always divisible by $3$, this is indeed the square of an integer.

Here's what I got from thinking about it for a little bit.

$u_1=16=1+5*10^0+10^1$
$u_2=1156=1+5*10^0+5*10^1+10^2+10^3$
$u_3=111556=1+5*10^0+5*10^1+5*10^2+10^3+10^4+10^5$
$u_k=1+\sum_{n=0}^{k-1} 5*10^n + \sum_{n=k}^{n=2k-1}10^n$
And $\sum_{n=k}^{n=2k-1}10^n=\sum_{n=0}^{n=2k-1}10^n-\sum_{n=0}^{n=k-1}10^n$

By the formula for the sum of a finite geometric series, we have: $$u_k=1+5 \frac{10^k-1}{9}+\frac{10^{2k}-1}{9} - \frac{10^k-1}{9}=1+ \frac{4(10^k)-4+10^{2k}-1}{9}$$ Bringing the 1 into the fraction and cancelling, we get$$u_k=\frac{10^{2k}+4(10^k)+4}{9}=\left(\frac{10^k+2}{3}\right)^2$$

And we are done.

$\rm\begin{eqnarray} {\bf Hint}\ & &\,\ 9\ (\overbrace{11\!\ldots\!11}^{\large \rm k}\overbrace{55\!\ldots\!556}^{\large\rm k}) \\[.2em] &= &\,\ 9\,(\color{#0a0}{11\!\ldots\ldots}\color{#0a0}{\ldots...\! 11} + \smash{\color{#c00}{\overbrace{44\!\ldots\!44}^{\large \rm k}}}\,+\,1) \\[.2em] &=&\rm\ \ \ \ \ \ \ \ \ \color{#0a0}{10^{\large 2k}-1}\ \ \ +\ \ \ \color{#c00}{4\,(10^{\large k} - 1)}\, +\, 9\\[.2em] &=&\rm\qquad\ \color{#0a0}{10^{\large 2k}} +\, \color{#c00}{4\cdot\!10^{\large k}} +\, 4 \\[.2em] &=&\rm\qquad (10{\large ^k}\ +\ 2)^{\large 2} \end{eqnarray}$