Prove that 16, 1156, 111556, 11115556, 1111155556… are squares. I'm 16 years old, and I'm studying for my exam maths coming this monday. In the chapter "sequences and series", there is this exercise:

Prove that a positive integer formed by $k$ times digit 1, followed by $(k-1)$
  times digit 5 and ending on one 6, is the square of an integer.

I'm not a native English speaker, so my translation of the exercise might be a bit crappy. What is says is that 16, 1156, 111556, 11115556, 1111155556, etc are all squares of integers. I'm supposed to prove that. I think my main problem is that I don't see the link between these numbers and sequences.
Of course, we assume we use a decimal numeral system (= base 10)
Can anyone point me in the right direction (or simply prove it, if it is difficult to give a hint without giving the whole evidence). I think it can't be that difficult, since I'm supposed to solve it.
For sure, by using the word "integer", I mean "natural number"  ($\in\mathbb{N}$)
Thanks in advance.

As TMM pointed out, the square roots are 4, 34, 334, 3334, 33334, etc...
This row is given by one of the following descriptions:


*

*$t_n = t_{n-1} + 3*10^{n-1}$

*$t_n = \lfloor\frac{1}{3}*10^{n}\rfloor + 1$

*$t_n = t_{n-1} * 10 - 6$


But, I still don't see any progress in my evidence. A human being can see in these numbers a system and can tell it will be correct for $k$ going to $\infty$. But this isn't enough for a mathematical evidence.
 A: Mark Bennet's hint seems to be a winner, so I'm reposting it CW:

Hint - try writing the general term in simple terms using the fact that a block of digits all the same can be summed as a geometric progression. So a sequence of $k$ '1's is $10^k−1\over9$, then see what you have. 

A: k = 1$\rightarrow$ $4^2 = 16$
k = 2$\rightarrow$ $34^2 = 1156$
k = 3$\rightarrow$ $334^2 = 111556$
k = 4$\rightarrow$ $3334^2 = 11115556$
etc
So,
the left part is given by: $(\frac{10^k - 1}{9} + 1)^2$
the right part is given by: $\frac{10^2k - 1}{9} + 4\frac{10^k - 1}{9} + 1$
work out both parts and you will see that they are equal. It is now proven since the base number of the left part (which is $\frac{10^k - 1}{9} + 1$) is always an integer.
A: Multiply one of these numbers by $9$, and you get $100...00400...004$, which is $100...002^2$.
A: 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.
A: Here's another way, using induction.  I'm not a huge fan of proofs by induction, as they often seem to me to mask what's going on.  In this case, though, induction allows you to stick to extremely elementary techniques, provided you can get your head around some notation and keep your columns in order.
Notation.  Let $3_{(k)}$ denote $k$-many 3's in a row. So, $3_{(2)} = 33$ and $10_{(2)}2 = 1002$, for instance.  
Our inductive hypothesis, following TMM's / Gerry Myerson's hint, is that $(3_{(n)}4)^2 = 1_{(n+1)}5_{(n)}6$.  Checking the base case is trivial: sure enough $(3_{(0)}4)^2 = 1_{(1)}5_{(0)}6$, i.e., $4^2 = 16$.
We now assume the result for $n=k$ and aim to prove it for $n=k+1$.  First, some elementary algebra, $(a+b)^2 = a^2 + 2ab + b^2$.   We use this as follows:
$(3_{(k+1)}4)^2 = (30_{(k+1)} + 3_{(k)}4)^2 = 90_{(2k+2)} + 20_{(k)}40_{(k+1)} + (3_{(k)}4)^2$ 
We then apply the inductive hypothesis to $(3_{(k)}4)^2$:
$= 90_{(2k+2)} + 20_{(k)}40_{(k+1)} + 1_{(k+1)}5_{(k)}6$ 
It's now a matter of adding the digits in each column.  A picture makes this easy:

In words: the last $k+1$ digits are just $5_{(k)}6$, as the left and middle terms have $0$'s here.  In the $k+2$th position, you have $4+1=5$, from the middle and last terms; so, cumulatively, we've now got $5_{(k+1)}6$.  For the next $k$ positions, the first and middle term again supply only zeroes.  So, we look to the final term, which gives use $k$ $1$'s.  We're now up to $1_{(k)}5_{(k+1)}6$.  At this point, the final term has run out and we look to the first and middle terms to fill the $2k+2$th column.  These give $9+2=11$.  So, we have $111_{(k)}5_{(k+1)}6$.  That is:
$(3_{(k+1)}4)^2 = 1_{(k+2)}5_{(k+1)}6$
So, we've proven the inductive hypothesis.
A: You may want to have a look at
http://www.cut-the-knot.org/arithmetic/NumberCuriosities/Squares.shtml
with a half dozen identities like that and links to as many of a similar sort. At one time there was a splash of activity on twitter.com with about the same question. But theis was in 2011.
A: 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.
A: $\rm\begin{eqnarray} {\bf Hint}\  & &\,\ 9\ (\overbrace{11\!\ldots\!11}^{\large \rm k}\overbrace{55\!\ldots\!556}^{\large\rm k}) \\[.2em]
&= &\,\ 9\,(\color{#0a0}{11\!\ldots\ldots}\color{#0a0}{\ldots...\! 11} + \smash{\color{#c00}{\overbrace{44\!\ldots\!44}^{\large \rm k}}}\,+\,1) \\[.2em]
&=&\rm\ \ \ \ \ \ \ \ \ \color{#0a0}{10^{\large 2k}-1}\ \ \ +\ \ \ \color{#c00}{4\,(10^{\large k} - 1)}\,  +\,  9\\[.2em]
&=&\rm\qquad\  \color{#0a0}{10^{\large 2k}} +\, \color{#c00}{4\cdot\!10^{\large k}} +\, 4 \\[.2em]
&=&\rm\qquad (10{\large ^k}\ +\ 2)^{\large 2} 
\end{eqnarray}$
A: Let $x=1\cdots1$ with $k$ digits.  Then your number is
$$\eqalign{x10^k+5x+1
  &=x(9x+1)+5x+1\cr
  &=9x^2+6x+1\cr
  &=(3x+1)^2\cr
  &=3\cdots34^2\ .\cr}$$
A: Here's my answer:
Let $a$ consists of $k$ digits, all equal to $1$.
$b=10...05$, where there are $(k-1)$ zeroes between $1$ and $5$.
Show that $ab+1$ is a positive integer for any positive value of $k$.
If $k=1$,
$a_1=1$, $b_1=15$, $a_1b_1+1=16$
If $k=2$,
$a_2=11$, $b_2=105$, $a_2b_2+1=1156$
If $k=3$,
$a_3=111$, $b_3=1005$, $a_3b_3+1=111556$
From this, we observe a geometric series between the terms.
If $k=m$,
$a_m= \frac{10^m-1}{9}$ and $b_m=10^m+5$.
Thus, we have
$a_mb_m+1= (\frac{10^m-1}{9}) \cdot (10^m+5) +1$
$= \frac{(10^m-1)(10^m+5)}{9} + \frac{9}{9}$
$= \frac{(10^{2m}-10^m+5\cdot10^m-5+9)}{9}$
$= \frac{(10^{2m}+4\cdot10^m+4)}{9}$
Since it can be written as a square, we have
$= (\frac{10^m+2}{3})^2$
But $k=m$,
$= (\frac{10^k+2}{3})^2$
From the previous answers, we already know that this is an integer.
