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 $n$.
 A: The formula would appear to be:
$n^2+(n^2+1)+\cdots+(n^2+n)=(n^2+n+1)+\cdots+(n^2+2n)$
Note that there are (n+1) terms on the left hand side and n terms on the right. As for a closed form of the sum, note that each side is an arithmetic series with a common difference of 1 and there are formulas you could use to compute this.
Note that the $n^2+2n+1=(n+1)^2$ for how the next row starts makes sense in the end as well.
An attempt at the closed form of the expression would be:
$n^2(n+1)+\frac{n(n+1)}{2}$
Where the first term is the $(n+1)$ occurances of $n^2$ whereas the second is the sum of the first $n$ natural numbers which is a well-known formula. If you want the sum of a row, you could just double this sum.
A: 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
$$
\begin{align}
n^2+n^3+\frac{n(n+1)}2
&=\frac{2n^3+3n^2+n}2\\
&=\frac{n(n+1)(2n+1)}2\tag{5}
\end{align}
$$
A: Both sides are $n (n+1)(2n+1)/2$ which also happens to be
$3 (1^2  + 2^2 + \ldots + n^2)$. 
A: $n^2 = \sum_{k=0}^{n-1} n = \sum_{k=0}^{n-1} ((n(n+2)-k)-(n(n+1)-k))$,
since $(n(n+2)-k)-(n(n+1)-k) = n$.
So we have
$n^2+\sum_{k=0}^{n-1} (n(n+1)-k) = \sum_{k=0}^{n-1} (n(n+2)-k)$.
A: 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.
A: Hopefully a proof without words.

A: Another way to build up the pattern you see.
Start with the $n$th triangle number, $T_n$ (I'll use $n=5$):
$$1 + 2 + 3 + 4 + 5 = 1 + 2 + 3 + 4 + 5.$$
Add $n^2$ to both sides, except distribute the $n^2$ on one side by adding $n$ to each of the $n$ terms:
$$5^2 + 1 + 2 + 3 + 4 + 5 = 6 + 7 + 8 + 9 + 10.$$
Finally, add $n^3$ to both sides by distributing $n^2$ to all terms except the first one on the left:
$$25 + 26 + 27 + 28 + 29 + 30 = 31 + 32 + 33 + 34 + 35.$$
So, both sides are
$$T_n + n^2 + n^3 = \frac{n(n+1)}{2} + n^2 + n^3.$$
