Easy№ 004 · Number theory · Geometry · posted by The Thinking Board

The sum of the first n odd numbers is a square

The problem

Show that for every positive integer $n$ ,
$1 + 3 + 5 + \dots + (2 n - 1) = n^{2} .$

This is the friendliest thing on the board, and it is here deliberately. It has a proof by induction, a proof by pairing, and a proof that is a picture with no words in it at all.

Comparing those three is the quickest way to see what this board is actually trying to measure. They are all correct. They are not all equally illuminating, and they are not equally easy to check — and those two virtues pull in opposite directions more often than people admit.

If you are posting here for the first time, post here.

Each L-shaped gnomon adds the next odd number — and keeps the square square

2 proofs posted

0 votes cast

Most convincing Newest

The picture, with as few words as I can manage

Board archive · 7 Sept 2026

To grow an $n \times n$ square into an $(n + 1) \times (n + 1)$ square, you lay an L-shaped strip along two of its sides: $n$ cells down one edge, $n$ along the other, and one in the corner. That is $2 n + 1$ cells.

The Greeks called that L a gnomon, after the shadow-arm of a sundial.

So the $k$ -th gnomon has $2 k - 1$ cells, and the first $n$ of them assemble — with nothing left over and nothing overlapping — into an $n \times n$ square:
$1 + 3 + 5 + \dots + (2 n - 1) = n^{2} .$

What the picture claims

That the identity is not a curiosity about odd numbers. It is what square means, taken apart. The odd numbers are exactly the gaps between consecutive squares,
$(k + 1)^{2} - k^{2} = 2 k + 1,$
so adding them up walks you from one square to the next by construction. There is nothing to verify once you have seen it; there is only something to look at.

This is among the oldest arguments in mathematics, and I would say it is still the best one on this page.

Each L-shaped gnomon adds the next odd number — and keeps the square square

Telescoping, which is the same proof with the picture removed

Board archive · 9 Sept 2026

Since $(k + 1)^{2} - k^{2} = 2 k + 1$ , the $k$ -th odd number is $k^{2} - (k - 1)^{2}$ . Sum from $k = 1$ to $n$ :
$k = 1 \sum n (2 k - 1) = k = 1 \sum n (k^{2} - (k - 1)^{2}) = n^{2} - 0^{2} = n^{2},$
since every interior term cancels the one after it.

A vote against my own post

This is airtight, it fits on one line, and it generalizes — the same trick sums cubes, fourth powers, or any polynomial you can write as a difference. By every practical standard it is the better proof.

It is also the gnomon proof with the drawing taken out.

I am posting it mainly to put a question to this board. If you find this more convincing than the picture, I suspect what you actually mean is that you find it more checkable — you can follow it symbol by symbol without trusting your eyes. That is a real virtue, and it is not the same virtue as explaining why the thing is true.

A proof can be easier to verify and harder to believe. Deciding which of those we are voting on seems worth settling early.