How to prove that $1+2=3, 4+5+6=7+8,... $ ad infinitum?

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!