Concept 6 of 19Foundation
Watch on YouTubeVideoPyramid (tetrahedral) numbers — triangles stacked
Imagine stacking oranges in a triangular-base pyramid:
- Layer 1: 1 orange
- Layer 2: 3 oranges (a triangle)
- Layer 3: 6 oranges
- Layer 4: 10 oranges
Total oranges in a pyramid of n layers = 1 + 3 + 6 + 10 + … + Tₙ.
This sequence is 1, 4, 10, 20, 35, 56, 84, 120, 165, 220, … — the tetrahedral (or pyramid) numbers.
Formula: Pₙ = n(n+1)(n+2)/6
Example
P₅ = 5·6·7/6 = 35.
P₁₀ = 10·11·12/6 = 220 — a cannonball pyramid 10 layers tall.
P₁₀ = 10·11·12/6 = 220 — a cannonball pyramid 10 layers tall.
💡 Tip:Tetrahedral = triangular-base pyramid. Don't confuse with square-base pyramids (those have different formula).
▸Why does this work? (derivation)
Why Pₙ = n(n+1)(n+2)/6?
Pₙ = T₁ + T₂ + … + Tₙ = Σ k(k+1)/2 = ½·[Σk² + Σk].
Using Σk = n(n+1)/2 and Σk² = n(n+1)(2n+1)/6:
Pₙ = ½·[n(n+1)(2n+1)/6 + n(n+1)/2]
= ½·n(n+1)·[(2n+1)/6 + 1/2]
= ½·n(n+1)·(2n+1+3)/6 = ½·n(n+1)·(2n+4)/6
= n(n+1)·(n+2)/6. ✓
Pₙ = T₁ + T₂ + … + Tₙ = Σ k(k+1)/2 = ½·[Σk² + Σk].
Using Σk = n(n+1)/2 and Σk² = n(n+1)(2n+1)/6:
Pₙ = ½·[n(n+1)(2n+1)/6 + n(n+1)/2]
= ½·n(n+1)·[(2n+1)/6 + 1/2]
= ½·n(n+1)·(2n+1+3)/6 = ½·n(n+1)·(2n+4)/6
= n(n+1)·(n+2)/6. ✓
🎯 Try it!
5 questions to check what you just read.
0 / 5
- Q1.P₄ (4th tetrahedral number)?
- Q2.Next: 1, 4, 10, 20, 35, ?
- Q3.Which is a tetrahedral number?
- Q4.P₁₀ (10-layer orange pyramid)?
- Q5.Formula for the nth tetrahedral number?