The sequence continues infinitely, why do the equations below work?
$$1+2=3$$ $$4+5+6=7+8$$ $$9+10+11+12=13+14+15$$
So I've been trying to observe some patterns but none seem to help me.
So I decided to expand this a bit more:
$$1+2=3$$ $$4+5+6=7+8$$ $$9+10+11+12=13+14+15$$ $$16+17+18+19+20=21+22+23+24$$ $$25+26+27+28+29+30=31+32+33+34+35$$
Does anyone want to offer any hints? I'm trying to figure out why they work
What's happening can be discovered by a scan, so I am hesitant to introduce symbols. But here goes. We want to show that $$\small n^2+(n^2+1)+(n^2+2)+\cdots +(n^2+n)=(n^2+n+1)+(n^2+n+2)+\cdots +(n^2+2n).\tag{1}$$ This can be done by using the usual formula for the sum of an arithmetic progression.
Maybe simpler is to note that the first term on the right of (1) is $n$ more than the second term on the left of (1). And the second term on the right of (1) is $n$ more than the third term on the left of (1), and so on.
So the sum of the $n$ terms on the right of (1) is $n^2$ more than the sum of the last $n$ terms on the left of (1). But the first term on the left of (1) is $n^2$. This completes the proof.
A pronic number is defined as $n(n+1)$ for $n\in\mathbb{Z^+}$.
We can write $n(n+1)-2\sum_\limits{i=1}^n=0$. And so each part of the sum can be split into $n(n+1)+i$ and $n(n+1)-i$, which upon re-arrangement gives your equations.
Also, you can see the first term on the LHS as $k^2$, with $k$ further sumends on the LHS, so just add $k$ to each one.
