From a Special Case to Every Case

Cases that fit

Add the odd numbers in order and watch the totals. 1 = 1, then 1 + 3 = 4, then 1 + 3 + 5 = 9, then 1 + 3 + 5 + 7 = 16. Every total so far is a square: $1^2$1², $2^2$2², $3^2$3² and $4^2$4².

Keep going and it keeps happening. The first five odd numbers add to $25=5^2$25 = 5², the first six to $36=6^2$36 = 6², and the first ten to $100=10^2$100 = 10².

Still a conjecture

The cases suggest a claim about every n: the first n odd numbers add to $n^2$n². Until an argument covers every n, it is a conjecture. Ten cases, or a million, leave infinitely many unchecked.

Patterns that hold for many cases can still fail. The number $n^2+n+41$n² + n + 41 is prime for every n from 0 to 39: 41, 43, 47, 53 and so on. At n = 40 it is 1600 + 40 + 41 = 1681, which is 41 × 41.

Mark some points on a circle and join every pair with a straight chord, placing the points so that no three chords meet at one point inside. With 1, 2, 3, 4 and 5 points the circle is cut into 1, 2, 4, 8 and 16 regions, doubling every time. With 6 points there are 31 regions, not 32.

A square in L-shaped layers

The argument comes from building the square. Start with 1 dot, a 1 by 1 square. Wrap an L-shaped layer of 3 dots round two of its sides and it becomes 2 by 2. Another layer of 5 makes it 3 by 3, and a layer of 7 makes it 4 by 4.

Each layer turns a k − 1 by k − 1 square into a k by k square, so the layer holds the difference: $k^2-{\left(k-1\right)}^2=k^2-\left(k^2-2k+1\right)=2k-1$k² − (k − 1)² = k² − (k² − 2k + 1) = 2k − 1. That is the kth odd number. Layer 1 holds 1, layer 2 holds 3, layer 3 holds 5, and layer 4 holds 7.

A 3 by 3 square. The 4 tinted dots are the first two layers, 1 + 3; the plain L of 5 dots is the third layer.

A 4 by 4 square. The 9 tinted dots are the first three layers, 1 + 3 + 5; the plain L of 7 dots is the fourth.

Every n at once

The layer count used nothing about a particular k. For every k, layer k holds 2k − 1 dots, and n layers build the n by n square. So $1+3+5+\cdots +\left(2n-1\right)=n^2$1 + 3 + 5 + … + (2n − 1) = n² for every n, and the conjecture is now a theorem.

Algebra gives the same result another way. The kth odd number is 2k − 1, so the first n of them add to 2(1 + 2 + … + n) − n. The sum 1 + 2 + … + n is $\frac{n\left(n+1\right)}{2}$n(n + 1)/2, so the total is $n\left(n+1\right)-n=n^2$n(n + 1) − n = n².

Once the result is proved, every particular case follows from it with no further work. The first 50 odd numbers add to $50^2=2500$50² = 2500, and the first 7 to 49, with nothing left to check.

What the special case assumed

A special case can hide an assumption. Suppose you check that if a > b then $a^2>b^2$a² > b², using a = 5 and b = 3: 25 > 9. Every case you try with positive numbers agrees, and a proof built from them may quietly use that a and b are positive.

The general claim is false: 1 > −2, yet $1^2=1$1² = 1 is less than ${\left(-2\right)}^2=4$(−2)² = 4. What was proved holds for positive numbers only. When an argument leans on a special case, ask what that case has that the general one lacks, such as a right angle, a positive sign or a whole number.

The usual mistakes

Calling a checked pattern proved. $n^2+n+41$n² + n + 41 passed 40 checks before failing.

Writing the claim with letters and stopping there. "The first n odd numbers add to $n^2$n²" states the claim for every n; it does not prove it.

Checking a proved result again for one case. Once the layers prove it for every n, the case n = 7 needs nothing more.

Proving a narrower claim than stated. An argument that works only for positive numbers proves the claim for positive numbers, whatever its first line says.

A training week and a spreadsheet

In the first application, two training plans suggest that a week’s total is 7 times the middle day, and naming the middle day m covers every plan at once. In the second, the first few entries of $n^3-n$n³ − n suggest 6 as the largest number dividing them all, and factoring proves it for every n.