Does this pattern of summing polygonal numbers to get a square repeat indefinitely?

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 $(n(n+3)/2)^2$.

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.