Geometric Bites

Proof by Induction: The Sum of Squares Formula Theorem. For any integer n ≥ 1, ∑ i=1 n i² = (1/6)·n·(n+1)·(2n+1) Proof (by mathematical induction) 1. Basis Step For n = 1: LHS = 1² = 1 RHS = (1/6)·1·(1+1)·(2·1+1) = (1/6)·1·2·3 = 1 Therefore, the formula holds for n = 1. 2. Inductive Hypothesis Assume the statement is true for some integer k ≥ 1: ∑(i=1→k) i² = (1/6)·k·(k+1)·(2k+1) 3. Inductive Step We must show it is true for n = k + 1: ∑(i=1→k+1) i² = [∑(i=1→k) i²] + (k+1)² = (1/6)·k·(k+1)·(2k+1) + (k+1)² = (1/6)(k+1)[k(2k+1) + 6(k+1)] = (1/6)(k+1)[2k² + 7k + 6] = (1/6)(k+1)(k+2)(2k+3) This matches the same form with k replaced by k+1: (1/6)·(k+1)·((k+1)+1)·(2(k+1)+1) Hence, the formula holds for n = k + 1. 4. Conclusion Since the statement is true for n = 1 (basis step), and true for n = k + 1 whenever it is true for n = k (inductive step), it follows by the principle of mathematical induction that ∑(i=1→n) i² = (1/6)·n·(n+1)·(2n+1) for al...

The Trapezoidal Rule — A Visual, First-Principles Introduction 11 November 2025 · @mathematics.proofs To understand integration deeply, it helps to think geometrically. Instead of memorising formulas, we begin by observing shapes. The goal is simple: break the interval into small pieces, estimate the area on each piece, and add everything together. Rectangles give a basic approximation. But functions rarely behave perfectly flat on every interval. A better idea is to allow the top edge to tilt. This leads us naturally to trapezoids . 1) Partitioning the Interval Consider an interval from a to b . We divide it into n equal parts. Δx = (b − a) / n x i = a + i·Δx for i = 0, 1, 2, …, n At each x i , we record the height of the function f(x i ). These sample values will guide our area estimates. 2) One Slice: Rectangle + Triangle Focus on a single subinterval [x i , x i+1 ]. If we draw a vertical line at x i and take a height of f(x i ), we obtain a rectangle o...