I’m cramming for a test and my brain keeps playing musical chairs with sequence patterns. I’m working on this one: 3, 8, 15, 24, 35, … and I want the next term and a proper nth-term rule, but I keep second-guessing myself.
Here’s my attempt so far: I wrote the first differences as 5, 7, 9, 11. Those go up by 2, so I figured it’s probably quadratic. I set a_n = a n^2 + b n + c and, starting at n = 1, I got:
– a + b + c = 3
– 4a + 2b + c = 8
– 9a + 3b + c = 15
Then I started wobbling about whether I should be starting at n = 0 instead (in which case c would equal the first term, right?), and somewhere in that indexing panic I solved it and got something that was off by 1 when I plugged it back in. Also, part of me thinks there’s a simpler story like “we’re adding consecutive odd numbers,” but I don’t know how to turn that into a clean formula without tripping.
Could someone walk me through a reliable, test-friendly way to spot the pattern here and set up the nth-term rule without messing up the indexing? And what’s a quick way to check that the rule actually matches all the given terms before I trust it to get the next one?
Follow-up question: if the first differences aren’t this neat, what’s your fast triage in a timed test-do you jump to ratios (geometric vibes), go straight to second differences, or look for alternating add/multiply behavior? Any tiny heuristic to keep me from spiraling would be amazing.
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3 Responses
You’re right to check differences: they’re 5, 7, 9, 11, so the second difference is 2 and the quadratic must have leading coefficient 1; solving from the first few terms gives a_n = (n+1)^2. Thus the next term is a_6 = 49; for example, a_3 = (3+1)^2 = 16.
Your sequence is doing the Odd-Number Conga: the jumps are 5, 7, 9, 11, so the second difference is a steady 2-classic “I’m a quadratic” vibes. A reliable test trick: if the constant second difference is d2, then the leading coefficient a is d2/2, so here a = 1. For a quadratic a_n = a n^2 + b n + c, the first difference is Δa_n = a(2n + 1) + b. Use n = 1 since Δa_1 = a_2 − a_1 = 8 − 3 = 5: that gives 3a + b = 5, so with a = 1 you get b = 2. Now plug a_1 = 3 into a + b + c = 3 to get c = 0. So a_n = n^2 + 2n = (n + 1)^2 − 1, which beautifully matches your “adding consecutive odds” story, because 3 + 5 + … + (2n + 1) = n^2 + 2n. Quick check: n = 3 gives 9 + 6 = 15, n = 5 gives 25 + 10 = 35, so we’re snug; hence the next term is a_6 = 36 + 12 = 48. Indexing tip: pick a starting index and don’t look back-n = 1 is perfectly fine here; if you start at n = 0 then c would equal a_0, not necessarily the first listed term. Speed-heuristics for tests: constant differences → linear, constant ratios → geometric, constant second differences → quadratic, alternating signs → sprinkle in (−1)^n, and if terms snuggle up to squares or cubes, compare to n^2 or n^3. Breathe, choose a lens, and the pattern usually stops playing musical chairs.
You were on exactly the right track with the first differences: 5, 7, 9, 11 means the second difference is constant (2), so think quadratic. My quick test trick is “a is half the second difference,” so here a = 1. Then use n starting at 1 (totally fine-starting at 0 also works, but I always trip over c when I do that): a_n = n^2 + b n + c. Plug n=1 and n=2: 1 + b + c = 3 and 4 + 2b + c = 8. Subtract to get b = 2, then c = 0. So a_n = n^2 + 2n = n(n+2). That also explains the “adding consecutive odd numbers” vibe: the terms are one less than perfect squares, 4−1, 9−1, 16−1, 25−1, … so the next term is 6^2 + 2·6 = 48. I did momentarily mis-remember and thought a might equal the second difference (so a = 2), but it’s actually half of it-glad I caught that before I embarrassed myself more than usual.
Fast check in a test: once you have a rule, generate the first three terms from it and make sure they match 3, 8, 15 (I sometimes also peek at the 4th for safety). For triage: if first differences are constant, it’s arithmetic; if ratios are constant, it’s geometric; if second differences are constant, it’s quadratic; and if third differences are constant, it’s cubic. If ratios are “almost” constant (like 2.0, 2.1, 2.0…), I still call that geometric-ish and then see if a small polynomial factor is hiding. Also scan for “near-square” shapes like (n+1)^2 − 1 or products like n(n+2). And if you do decide to start at n=0, then yes, c equals the first term-though I weirdly still write the first data point as n=1 out of habit, which is probably not best practice but keeps me from spiraling.