Arithmetic pattern $1 + 2 = 3$, $4 + 5 + 6 = 7 + 8$, and so on I am noticing this pattern:
\begin{align}
1+2&=3\\
4+5+6&=7+8\\
9+10+11+12&=13+14+15 \\
16+17+18+19+20&=21+22+23+24 \\
&\vdots
\end{align}
Is there a general formula that expresses a pattern here, such as for the $n$th term, or the closed form of the entire summation? I think the $n$th term starts with $$n^2+(n^2+1)+\cdots=\cdots+[(n+1)^2-1].$$
I am also presuming once this formula is discovered, we can prove it by induction for any $n$.

Proof of the Identities
$$
\begin{align}
n^2&=\sum_{k=1}^nn\tag{1}\\
n^2+\sum_{k=1}^n\left(n^2+k\right)&=\sum_{k=1}^n\left(n^2+n+k\right)\tag{2}\\
\sum_{k=0}^n\left(n^2+k\right)&=\sum_{k=n+1}^{2n}\left(n^2+k\right)\tag{3}
\end{align}
$$
Explanation:
$(1)$: multiplication written as a sum
$(2)$: add $\sum\limits_{k=1}^n\left(n^2+k\right)$ to both sides
$(3)$: include $n^2$ in the sum on the left and reindex the sum on the right

Derivation of the Formula for Each Side
Using the formula
$$
\sum_{k=1}^nk=\frac{n(n+1)}2\tag{4}
$$
and equation $(2)$, we can compute the sum of each side of $(3)$ as
$$
\begin{align}
n^2+n^3+\frac{n(n+1)}2
&=\frac{2n^3+3n^2+n}2\\
&=\frac{n(n+1)(2n+1)}2\tag{5}
\end{align}
$$
