A tile pattern that grows the SAME amount every time. We build to Design 5 and pull out the equation. Click Next.
Green = the starter tile (never changes). Blue = old tiles. Orange = 3 NEW tiles just added.
The jump is +3 every time. Same jump = linear. No second subtraction needed.
Take paper. Draw them. Count the tiles. Then click Next.
Two clues: the jump, and the starter tile.
Clue 1 — the JUMP is the MULTIPLIER.
Each design adds +3 → start with 3 × n
Test it: Design 1 → 3 × 1 = 3. But Design 1 has 4 tiles!
One tile is missing… who is it? 🔍
Clue 2 — the missing tile is the green STARTER. It is there before any columns — even "Design 0" has it. So add + 1.
tiles = 3 × n + 1
3 = the jump. n = design number. +1 = the starter tile.
Test it on a design we counted. Then jump far away.
Test on Design 3 (we counted 10):
3 × 3 + 1 = 9 + 1 = 10 ✓ Design 5: 3 × 5 + 1 = 16 ✓
Design 100 — no drawing needed:
3 × 100 + 1 = 300 + 1 = 301 tiles
Do NOT multiply the first design. Design 1 = 4, so many students write tiles = 4 × n. Test it: Design 2 → 4 × 2 = 8. Wrong! (Real answer: 7.) The multiplier is the JUMP (+3), not the first number.
Second trap: forgetting the + 1 starter. Always test your equation on TWO designs you can count.
The jump is the multiplier. The starter is the plus. tiles = jump × n + starter.
A tile pattern grows: Design 1 has 5 tiles, Design 2 has 9, Design 3 has 13. Which equation gives the number of tiles in Design n?
Answer: (A) t = 4n + 1.
1. Jumps: +4, +4 → same → linear. The multiplier is 4.
2. Starter: Design 1 = 5, take away one jump → 5 − 4 = 1. So t = 4n + 1.
3. Trap check: (B) t = 5n gives 5 for n=1 ✓ but 10 for n=2 ✗. One design is not enough — always test TWO!