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 $n$.

Let's examine a simple identity:
$$9+10+11+12=13+14+15$$
Notice that there are $4$ terms on the left and $3$ on the right - so let's "use up" one of the terms on the left. We can split $9=3+3+3$ and distribute these additional $+3$ terms to the other $3$ summands on that side to get:
$$(10+\color{red} 3)+(11+\color{red} 3)+(12+\color{red} 3)=13+14+15$$
and, of course, the left hand is exactly $13+14+15$ when we compute. Well, that was easy!
Let's generalize. The sum is always of the form
$$n^2+(n^2+1)\ldots+(n^2+n)=(n^2+n+1)+\ldots + (n^2+2n)$$
and there are $n+1$ terms on the left hand and $n$ terms on the right hand side. So how do we get from left to right? Easy! We take the $n^2$ term and divide it into $\underbrace{n+n+\ldots+n}_{n\text{ times}}$. Now, there are $n$ other summands on that side, so we distribute a $+n$ term to each of them without changing the value. This yields
$$(n^2+1+\color{red} n)+\ldots+(n^2+n+\color{red}n)=(n^2+n+1)+\ldots + (n^2+2n)$$
but this is obvious, since simplifying both sides gives
$$(n^2+n+1)+\ldots+(n^2+2n)=(n^2+n+1)+\ldots + (n^2+2n)$$
where both sides are identical.
