Picture a clock whose rim gets a little bigger every time the hand goes all the way around. Follow the tip of the hand: each rotation it travels one full circumference, and each circumference is longer than the last by the same amount. So the distance per rotation climbs in a straight line, but the total distance bends upward into a curve. In this lesson you will see why by experimenting, then work out the formula step by step, then prove it.
Where this lesson is headed
In words: total distance = (rotations × starting circumference) + (growth × the number of growth steps that have piled up). You do not need to memorize this. By the end you will be able to rebuild it yourself and explain why it is true.
N: starting circumference (how far the tip travels in the first rotation)
g: how much the circumference grows after each rotation
k: number of rotations completed
T(k): 0 + 1 + ⋯ + (k − 1), the growth steps piled up over k rotations
S(k): total distance after k rotations
The clock starts with circumference N = 1 cm, and the rim grows by g = 1 cm after every full rotation. So the tip of the hand travels 1, 2, 3, 4, … cm on successive rotations. Turn the hand a few times and watch the total. Then change N or g: your rotation count stays put, so you can see exactly how the total changes. The guided experiments below walk you through it. (Shortcut: Space turns one rotation, ← undoes one.)
The gold dot is the tip of the hand. It rides along the rim, and the distance it travels is what we measure. The drawing widens when the rim grows, but not to scale.
What will the total distance be after 1 rotation?
Changing a setting keeps your rotation count, so you can compare totals directly.
Do these in order. Each one changes a single thing so you can see what that setting does. The "Set this up" button sets the clock for you; the drawing, graph and table update with it.
Rotation j is the hand's j-th trip around. Before it starts, j − 1 rotations are already finished, and each of those grew the rim by g.
| Rotation j | Distance this rotation N + (j − 1)g | Running total S(j) |
|---|
Each step below has two sides: the general rule on the left, and the same step done with actual numbers on the right. The numbers use your settings from Stage 01. Start with the simplest clock, then try other settings and watch every step still work.
Working example: N = 1, g = 1
The first rotation covers N. Every finished rotation grows the rim by g. When rotation j starts, j − 1 rotations are finished, so the rim has grown j − 1 times:
The "− 1" matters: rotation 1 has had no growth yet, so L(1) = N.
The total is every rotation added together:
Each rotation is "N plus some growth". Collect all the N's in one pile and all the growth in another. There are k copies of N, and the growth amounts are 0g, 1g, 2g, …, (k − 1)g:
In the picture, each bar is one rotation. The blue pieces form a rectangle (k × N). The gold pieces form a staircase (the growth).
Call the sum A. Write it forwards, then write it again backwards underneath. Each column adds to the same number, k − 1 (small numbers on top meet big numbers below). There are k columns, so the two copies together make k(k − 1):
In the step 2 picture, this is the same as fitting two copies of the gold staircase together into a k by (k − 1) rectangle and taking half. We call it T(k); it starts at 0, which is why it has k − 1 and not k + 1.
Replace the sum in step 2 with the shortcut from step 3:
This is the formula at the top of the page. The first part is the "no growth" straight line from Stage 01; the second part is the extra from growth.
Distribute and collect terms:
A formula with a k² term is quadratic, and its graph is a parabola. The k² comes only from the growth, with coefficient g/2. If g = 0 the k² term disappears and the total is a straight line.
Look at how much the total goes up from one rotation count to the next. That jump is just the distance of the next rotation, N + kg. Now look at how much those jumps change: always exactly g.
When the "change of the change" is constant, the pattern is quadratic. (For a straight line, the change itself is constant.) A quadratic ak² + bk has change-of-change 2a, so here 2a = g and a = g/2, matching step 5.
A good formula should give obvious answers in obvious cases:
k = 0: no rotations, so S(0) should be 0.
k = 1: one rotation covering N, so S(1) should be N.
g = 0: every rotation covers N, so S(k) should be kN.
N = g: then S(k) = N · k(k + 1)/2, a multiple of the familiar 1, 3, 6, 10, 15, ….
Same N, g and k as above.
Method A · add every rotation
Method B · use the formula
Two matching answers confirm this one case. They do not show the formula works for every k. The steps above explain why it works in general, and Stage 03 proves it.
We will prove the identity for every nonnegative integer k, for any fixed starting circumference N and growth g. Induction works because the next total is obtained by adding precisely one new rotation.
Complete proof
Base case. S(0) = 0, while 0·N + g·0(0 − 1)/2 = 0.
Induction hypothesis. Suppose S(k) = kN + gk(k − 1)/2 for some k ≥ 0.
Next rotation. Rotation k + 1 has circumference N + kg. Hence
This is the original formula with k replaced by k + 1. Therefore it holds for every nonnegative integer k. ∎