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 term in the left side (which is equal to $n^2$) may be redistributed among the remaining $n$ terms to increase each one of them by $n$.

First you have to show (by induction or otherwise) that the summation limits for the $n$th line are actually the following:

$$\sum_{i=n^2}^{n^2+n} i=\sum_{i=n^2+n+1}^{n^2+2n}i$$

This equation can then be shown to be true just by working out what those sums are and simplifying.

Here is a method that doesn't explicitly use induction.

For any positive integer $n$, we want to show that $$\sum_{k=n^2}^{n^2+n} k=\sum_{k=n^2+n+1}^{n^2+2n} k$$.

The left hand sum has $n+1$ terms, each with a common sub-term of $n^2$. Thus the left hand sum can be rewritten as $${\rm{LHS}}= (n+1)\cdot n^2 + 0 + 1+\cdots+n$$

The right hand sum has $n$ terms, each with a common sub-term of $n^2+n$, and so can be rewritten as $${\rm{RHS}}=n\cdot(n^2+n)+1+2+\cdots+n$$

You can probably see how to finish up!

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: $4,5,6,\cdots,13,14,15$.