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 $n+1$

• It's just summing a bunch of arithmetic progressions. Commented Aug 28, 2017 at 14:56
• Commented Aug 28, 2017 at 15:01
• Are you specifically looking to prove this using induction? If so, then please make that clear in the question. Commented Aug 28, 2017 at 15:02
• I feel there will be a nice geometric interpretation using triangular numbers for a picture style proof.
– Karl
Commented Aug 28, 2017 at 15:09
• I love this kind of relations. Here is another spectacular one $$\begin{array}{*{20}{r}} 1& = &1 \\ {3 + 5}& = &8 \\ {7 + 9 + 11}& = &{27} \\ {13 + 15 + 17 + 19}& = &{64} \\ {21 + 23 + 25 + 27 + 29}& = &{125} \\ \ldots & \ldots & \ldots \end{array}$$ Commented Aug 28, 2017 at 16:50

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 $n$.

• I think this leads to the same argument as that made by @paw88789. My problem was that I wrote the problem in a too complicated way...
– Arne
Commented Aug 28, 2017 at 20:13
• @Arne Simplifying is an art. I feel lucky that I spotted this one. Commented Aug 28, 2017 at 21:12

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!

• Ok, after factoring out $n$ in RHS this shows that both sides are equal. Thanks. This makes the induction step trivial...
– Arne
Commented Aug 28, 2017 at 20:17

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

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

$$n^2+n=n(n+1)$$

$$n^2+(n^2+1)\ldots+(n^2+n)=[n^2+(n+1)]+[(n^2+1)+(n+1)]+\ldots+[(n^2+n-1)+(n+1)]$$

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: $$\sum_{i=k}^{k+m}i = \sum_{i=k+m+1}^{k+2m}i \Leftrightarrow k = m^2$$