The picture behind the formula (Fig. 8.5). Red circles count 1 + 2 + 3 + 4 + 5 + 6 = 21; green circles count the same again; together they fill a 7 × 6 rectangle.
Why it happens: two copies of the staircase 1, 2, 3, …, n interlock into a rectangle n + 1 wide and n tall. That rectangle holds n(n + 1) circles, and it is exactly twice the sum, so Sn = n(n + 1)/2. Because the argument is about how the two staircases fit, it holds for n = 20 and n = 1000 just as it does for the n = 6 you can draw.
Check it yourself: one of n and n + 1 is always even, so n(n + 1)/2 is always a whole number — as a count of circles must be.
Q2.
Let us revisit the sequence tn of triangular numbers 1, 3, 6, 10, 15, … shown in Fig. 8.1. Note that the nth term of this sequence is the sum of the first n natural numbers. Thus tn = n(n + 1)/2. Can you use this to find the 10th, 17th and 80th triangular numbers?
Check the first of these directly: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55. ✓
Why it happens: the two ideas in this section are one idea. A triangular array with n rows has 1 dot in the top row, 2 in the next, and so on down to n — so counting its dots is adding the first n natural numbers. Once you know that sum is n(n + 1)/2, you know every triangular number without drawing a single dot. Drawing the 80th array would need 3240 dots; the formula needs one multiplication and one division.
Try This: add two consecutive triangular numbers, say 55 + 66. You get 121 = 112. Two triangles always fit together into a square.