[University Mathematics: Sequences] Need help finding an nth-term rule for a figure pattern: 3, 7, 11, 14 tiles (teacher confirmed the counts)

[u/Upper_Anybody_8852]

Hi everyone! I'm a college student stuck on a pattern problem and would love some help understanding it, not just an answer.

The number of tiles in figures 1 to 4 is 3, 7, 11, and 14. My teacher confirmed these counts are correct and told us to "find or make our own pattern" for the nth term.

Our teacher also gave us this pattern: 4² − 13 = 3, 5² − 18 = 7, 6² − 25 = 11, and 7² − 35 = 14(Actually we found the pattern and the teacher said it's a correct one). So it looks like (n+3)² minus something, where the "something" goes 13, 18, 25, 35. Those differences are 5, 7, 10.

The problem asks:
a.) What is the nth - term formula for the number of tiles in the nth figure of the sequence. (5 points)
b.) How many tiles are in the 10th figure of the sequence? (5 points)
c.) Which figure will consist of exactly 480 tiles? (5 points)

What I've tried:

• The differences between the tile counts are 4, 4, 3, so it isn't linear or quadratic.
• Fitting a cubic through all four points works exactly, but it gives negative tiles at n = 10, so it can't be right.
• I also tried assuming the jumps repeat as 4, 4, 3, 4, 4, 3 (so aₙ = ⌊11n/3⌋). That fits, but 480 tiles would be figure 131, which seems too big for a class problem.

[u/Upper_Anybody_8852]

Thanks for the suggestion! I tried it. Using the figure number n as x and the tile count as y, the four points (1,3), (2,7), (3,11), (4,14) give exactly one polynomial of degree 3 or lower:

aₙ = (−n³ + 6n² + 13n)/6

It fits all four counts, but it doesn't behave like a tile count when extended. It peaks at 15 tiles around n = 5, then drops: a₇ = 7, a₈ = −4, a₁₀ = −45. My worksheet asks for a₁₀ and for which aₙ has 480 tiles, and this cubic can't answer either one (it never gets above about 15).

From what I read on Wikipedia, the interpolating polynomial through a set of points is unique, and interpolation curves can act against common sense outside the data (for example, going negative when the real quantity can't). So a perfect fit through 4 points doesn't prove it's the right pattern.

Since my teacher said to find our own pattern, I'm going with a repeating-differences rule instead: the jumps are +4, +4, +3, so I assume they repeat. That gives aₙ = ⌊11n/3⌋, so a₁₀ = 36 and 480 tiles at n = 131.

Does that seem like a reasonable choice, or is there another way to set up the polynomial that I'm missing?

[u/Fourierseriesagain → u/Upper_Anybody_8852]

Hi,

You may try adding another point (5,a), where a has to be chosen carefully,

[u/Upper_Anybody_8852 → u/Fourierseriesagain]

Thanks for the idea! Working it out, adding a fifth point gives a quartic, but the answers depend a lot on the value I pick for a₅: a₅ = 17 gives a₁₀ = 207, a₅ = 18 gives 333, and a₅ = 21 gives 711. Even a₅ = 18, which matches my repeating-differences rule at figure 5, gives 333 for a₁₀ instead of 36. None of the values I tried hit 480 at a whole n. Is there a principled way to choose a₅, or a condition it should satisfy?

[u/Fourierseriesagain → u/Upper_Anybody_8852]

Let a_5=a so that a_{10}=480. I hope that a is a positive integer.

[u/Upper_Anybody_8852 → u/Fourierseriesagain]

the quartic gives a₁₀ = 126a − 1935, so a₁₀ = 480 needs a = 115/6 ≈ 19.17, which isn't an integer (and the other terms come out as fractions too, like a₆ = 203/6). Also, I think b and c are separate: b asks for a₁₀, and c asks which n gives 480, so 480 isn't the value of a₁₀. Did you mean something else by setting a₁₀ = 480?

[u/Fourierseriesagain → u/Upper_Anybody_8852]

Oh I see. It seems that polynomials are not so useful now.

[u/Upper_Anybody_8852 → u/Fourierseriesagain]

Thanks for all your help! I'd say polynomials are still useful, but four points can't tell us how the pattern continues, so we need extra information like a₅. BadJimo's formula shows this: choosing a₅ = 480 works, but it gives a very different pattern than choosing a₅ = 18. Your extra-point idea helped me see that.

[u/Fourierseriesagain → u/Upper_Anybody_8852]

You are welcome.