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 $n+1$
 A: 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.
A: 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!
A: 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: $4,5,6,\cdots,13,14,15$.
A: 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 $n$.
A: Suppose there are $n$ terms on the Right-Hand side. Then there are $n+1$ terms on the Left-Hand Side. Notice the last term of the Left-Hand Side can be evenly 
distributed to the other $n$ terms, transforming them into terms on the Right-Hand Side.
To be specific
$$n^2+n=n(n+1)$$
$$n^2+(n^2+1)\ldots+(n^2+n)=[n^2+(n+1)]+[(n^2+1)+(n+1)]+\ldots+[(n^2+n-1)+(n+1)]$$
A: So, what you want to do is:
$$\{k,m\} \in \mathbb{N} \times \mathbb{N}:\sum_{i=k}^{k+m}i = \sum_{i=k+m+1}^{k+2m}i$$
(I have corrected the upper limit of the first sum.)
Let note $S_n$ the sum of the $n$ first naturals: $S_n = \sum_{i=0}^{n}i = \frac{n(n+1)}{2}$.
If you explicit each line of the set of equations, you find that:
\begin{align}
1 + 2 = 3 &\Leftrightarrow S_2 - S_0 = S_3 - S_2 \\
4 + 5 + 6 = 7 + 8 &\Leftrightarrow S_6 - S_3 = S_8 - S_6 \\
9 + 10 + 11 + 12 = 13 + 14 + 15 &\Leftrightarrow S_{12} - S_8 = S_{15} - S_{12} \\
&\vdots \\
\sum_{i=k}^{k+m}i = \sum_{i=k+m+1}^{k+2m}i &\Leftrightarrow S_{k+m} - S_{k-1} = S_{k+2m} - S_{k+m} \\
&\Leftrightarrow 2S_{k+m} - S_{k-1} = S_{k+2m}
\end{align}
Then, if you replace $S_n$ by its formula in the right-side equation and derive it, you get:
$$\sum_{i=k}^{k+m}i = \sum_{i=k+m+1}^{k+2m}i \Leftrightarrow k = m^2$$
