How to prove that $1+2=3, 4+5+6=7+8,... $ ad infinitum? Given this set of equations:
$$
1+2=3\\
4+5+6=7+8\\
9+10+11+12=13+14+15\\
\ldots
$$
How can I prove that this is true for all continuations of this sequence?
I would put it in the form of:
$$
(k,m)\in \{n^2,n|\in\Bbb N\}\\
\sum_{i=k}^{k+m} i=\sum_{i=k+m+1}^{k+2m}i
$$
However, I have problems in formulating and solving the inductions step, which I think should be to go from $n$ to $n+1$

The first row is true:
$$1+2=3.$$
Swap the sides of the first row and add to the second row to get:
$$3+4+5+6=1+2+7+8.$$
This is true, because in the arithmetic progression $1,2,3,4,5,6,7,8$, the sums of terms equidistant from the center are equal.
Similarly, swap the sides of row $2$ and add to row $3$:
$$7+8+9+10+11+12=4+5+6+13+14+15.$$
Again the sum of the central terms is equal to the sum of the external terms in the AP: $4,5,6,\cdots,13,14,15$.
