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.
 A: Let the first term be a triangular number $n(n+1)/2$, the common difference between successive terms be $n(n-1)/2$, and the number of terms be $n+3$.  Then the average of all terms is
$\dfrac{n(n+1)}{2}+\dfrac{n+2}{2}×\dfrac{n(n-1)}{2}=\dfrac{n^2(n+3)}{4}$
Multiplying this by the number of terms $n+3$ then gives a sum of $(n(n+3)/2)^2$.
A: This is a much more drawn out approach using summations.  It doesn't have the elegance of Oscar's solutions, but it still works.  You have that with $T_k$ being the $k$-th Triangular number (here $T_0=0$)
$$\sum_{k=1}^{4+0}{[T_1+(k-1)T_0]}=(T_1+1)^2$$
$$\sum_{k=1}^{4+1}{[T_2+(k-1)T_1]}=(T_2+2)^2$$
$$\sum_{k=0}^{4+2}{[T_3+(k-1)T_2]}=(T_3+3)^2$$
This suggests that 
$$\sum_{k=1}^{m+4}{[T_{m+1}+(k-1)T_{m}]}=(T_{m+1}+(m+1))^2$$
Since summation is linear, we have
\begin{eqnarray*}\sum_{k=1}^{m+4}{[T_m+(k-1)kT_{m-1}]}&=&T_{m+1}\sum_{k=1}^{m+4}1+T_m\sum_{k=1}^{m+4}{(k-1)}\\&=&T_{m+1}(m+4)+T_m\sum_{k=1}^{m+3}{k}
\end{eqnarray*}
Using the formula for the $m$-th Triangular number, we have
$$=\left[\frac{(m+1)(m+2)}{2}\right](m+4)+\left[\frac{m(m+1)}{2}\right]\left[\frac{(m+3)(m+4)}{2}\right]$$
Factoring and simplifying gives 
$$\frac{(m+1)(m+4)}{2}\left[(m+2)+\frac{m(m+3)}{2}\right]=\frac{(m+1)^2(m+4)^2}{4}$$
Now, the right hand side
\begin{eqnarray*}[T_{m+1}+(m+1)]^2&=&T_{m+1}^2+2(m+1)T_{m+1}+(m+1)^2\\&=&\frac{(m+1)^2(m+2)^2}{4}+2(m+1)^2\frac{(m+2)}{2}+(m+1)^2
\end{eqnarray*}
Factoring out an $(m+1)^2$,
\begin{eqnarray*}\frac{(m+1)^2(m+2)^2}{4}+2(m+1)^2\frac{(m+2)}{2}+(m+1)^2&=&(m+1)^2\left[\frac{(m+2)^2}{4}+\frac{4(m+2)}{4}+\frac{4}{4}\right]\\&=&\frac{(m+1)^2}{4}\left[(m^2+4m+4)+(4m+8)+4\right]\\&=&\frac{(m+1)^2}{4}(m^2+8m+16)\\&=&\frac{(m+1)^2(m+4)^2}{4}
\end{eqnarray*}
Since both expressions equal $\frac{(m+1)^2(m+4)^2}{4}$, we are done.
EDIT:
For the new problem, note the first term in each line is a successive square and as above, the common difference is a triangular number.  Thus we have to prove, in general, 
$$\sum_{k=1}^n[n^2+(k-1)T_{n-1}]=T_n^2$$
The left hand side then can be written as 
\begin{eqnarray*}\sum_{k=1}^n[n^2+(k-1)T_{n-1}]&=&\sum_{k=1}^n n^2+T_{n-1}\sum_{k=1}^n(k-1)\\&=&n^3+T_{n-1}\sum_{k=1}^{n-1}k\\&=&n^3+\frac{(n-1)n}{2}\cdot\frac{(n-1)n}{2}\\&=&n^2\left(n+\frac{(n-1)^2}{2^2}\right)\\&=&\frac{n^2}{2^2}(n+1)^2\\&=&\left[\frac{n(n+1)}{2}\right]^2\\&=&T_n^2
\end{eqnarray*}
And we're done.
A: 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*}
