Does this pattern of summing polygonal numbers to get a square repeat indefinitely? I am using the table of polygonal numbers on this site:   
http://oeis.org/wiki/Polygonal_numbers 
The first row of the table gives the value of $n=0,1,2,3,...$. The first column gives the polygonal numbers $P_{N}(n)$ starting with $N=3$.
Here's the pattern, the sum is done by adding elements of the same column starting with the triangular numbers.
$$1+1+1+1=4=2^2$$
$$3+4+5+6+7=25=5^2$$
$$6+9+12+15+18+21=81=9^2$$
$$10+16+22+28+34+40+46=196=14^2$$
...
The squares on the right hand side are given by $n+T_{n}$. The sum of index $(n+1)$ has one more element than that of $n$.
1-Does this pattern repeat indefinitely? (I suspect the answer is yes but I can't prove it)
2-Do we know why we have such a pattern?
Edit 03-05-2019
Following the suggestion of Eleven-Eleven, I looked for other patterns similar to the one above. I found one that is even simpler. This time we skip the triangular numbers when we calculate the sum. We start with the squares and sum up enough terms to get another square.
$$1=1^2=T_{1}^2$$
$$4+5=3^2=9=T_{2}^2$$
$$9+12+15=6^2=36=T_{3}^2$$
$$16+22+28+34=10^2=100=T_{4}^2$$
$$25+35+45+55+65=15^2=225=T_{5}^2$$
We see the same pattern as above. The square with index $(n+1)$ requires the addition of one more term than the square with index $n$. The number of elements to sum up to get the square $T_{n}^2$ is simply $n$.  
Can this be stated as the following theorem?  
The square of a triangular number $T_{n}$ can be expressed as the sum of $n$ polygonal numbers excluding the triangular number itself.

I'm adding another answer because I wanted to try and use the formula for polygonal numbers as given here in an attempt to better answer the original question by the OP.  In both cases above we exploited the arithmetical sequence properties and only used Triangular numbers.  Can we get the same result strictly using the polygonal number formula
$$P(s,n)=\frac{n^2(s-2)-n(s-4)}{2}$$
where $n$ represents the sequence number and $s$ represents the number of sides in a polygonal number.  So, lets look at an example.  In the second row, the OP has that 
$$3+4+5+6+7=25=(T_2+2)^2$$
Well, $3,4,5,6,$ and $7$ are the second triangular, square, pentagonal, hexagonal, and heptagonal numbers, so $n=2$ and $s$ is indexed... therefore, we have
\begin{eqnarray*}\sum_{k=1}^5{P(k+2,2)}&=&\sum_{k=1}^5{\left[\frac{4((k+2)-2)-2((k+2)-4)}{2}\right]}\\&=&\sum_{k=1}^5{\left[\frac{4k-2(k-2)}{2}\right]}\\&=&\sum_{k=1}^5\left(k+2\right)\\&=&\sum_{k=1}^5{k}\\&=&T_5
\end{eqnarray*}
Now, this can be generalized.  The claim is
$$\sum_{k=1}^{m+3}{P((k+2),m)}=\left(T_m+m\right)^2$$
Proof:
\begin{eqnarray*}\sum_{k=1}^{m+3}{P((k+2),m)}&=&\sum_{k=1}^{m+3}\left[\frac{m^2((k+2)-2)-m((k+2)-4)}{2}\right]\\&=&\sum_{k=1}^{m+3}\left[\frac{m^2k-mk+2m}{2}\right]\\&=&\frac{m^2}{2}\sum_{k=1}^{m+3}{k}-\frac{m}{2}\sum_{k=1}^{m+3}{k}+m\sum_{k=1}^{m+3}1\\&=&\frac{(m-1)m}{2}\sum_{k=1}^{m+3}k+m(m+3)\\&=&\frac{(m-1)m}{2}\frac{(m+3)(m+4)}{2}+\frac{4m(m+3)}{4}\\&=&\frac{m(m+3)}{4}\left[(m-1)(m+4)+4\right]\\&=&\frac{m(m+3)}{4}\left(m^2+3m\right)\\&=&\frac{m^2(m+3)^2}{2^2}\\&=&\left[\frac{m(m+3)}{2}\right]^2\\&=&\left[\frac{m(m+1+2)}{2}\right]^2\\&=&\left[\frac{m(m+1)+2m}{2}\right]^2\\&=&\left(T_m+m\right)^2
\end{eqnarray*}
