I’m preparing for a test and I’m stuck on a reliable way to get the nth term when the sequence isn’t arithmetic. For example: 2, 5, 10, 17, 26, …
My first try was a linear rule an + b. Using the first two terms I got 3n − 1, which matches n=1 and n=2 but then it fails at n=3 (gives 8 instead of 10). So I checked differences: 3, 5, 7, 9, which gives a constant second difference of 2. I remember that means it should be quadratic and that the coefficient of n^2 is half the second difference, so a=1. Then I wrote T(n) = n^2 + bn + c and tried solving for b and c using the first couple of terms. This is where I keep tripping: I’m not sure whether to index from n=1 or n=0, and depending on what I pick I get different b and c. Once I got b=0, c=1, another time I ended up with b=−1, c=2, and I’m not convinced either approach is consistent.
An analogy that helps me (maybe wrongly) is thinking of it like building rows of tiles where each new row has two more tiles than the previous one-the total feels like “summing odd numbers,” which I recall links to square numbers. But I don’t see the clean step from that idea to the exact nth-term rule without making the same algebra slips.
What is the most straightforward, test-friendly method to go from the second-difference table to the nth-term rule, including how to choose the starting index (n=0 vs n=1) so I don’t tie myself in knots? Also, if the sequence is like “a square pattern plus a constant” or starts later (e.g., data corresponds to n starting at 3), how should I adjust the rule systematically without guessing?
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3 Responses
I use a little “differences recipe” so I don’t have to solve simultaneous equations each time. If the second difference is constant, the rule is quadratic: T(n) = an^2 + bn + c. Work with n starting at 1 (that’s what most tests assume). Then: a = (second difference)/2; b = ΔT(1) − 3a, where ΔT(1) = T(2) − T(1); and c = T(1) − a − b. Worked example on your sequence 2, 5, 10, 17, 26: second differences are 2, so a = 1; ΔT(1) = 5 − 2 = 3, so b = 3 − 3·1 = 0; then c = 2 − 1 − 0 = 1. Hence T(n) = n^2 + 1, which matches 1→2, 2→5, 3→10, etc. This also lines up with your “sum of odds gives squares” intuition: you’re basically at the square numbers with a +1 shift. If your table is indexed differently, you can either reindex so the first given term is n = 1 (set m = n − k + 1 if your first data point is at n = k, find T in terms of m, then replace m by n − k + 1), or use the base-k version of the same recipe: a = (second difference)/2, b = ΔT(k) − a(2k + 1), c = T(k) − a k^2 − b k. Either way, it’s systematic and stops the “am I starting at 0 or 1?” spiral-I’ve been there!
Take half the second difference for the quadratic coefficient a, subtract a·n^2 from the sequence to “flatten” it, then fit what’s left linearly with your chosen index (I default to n=1 in exams): here a=1, the remainders are all 1, so T(n)=n^2+1. I used to tie myself in indexing knots until a kind teacher showed me the shift-safe version T(n0+k)=T(n0)+ΔT(n0)·k+(Δ^2T(n0)/2)·k(k−1), which lets you start at any n0 (like 3) and build the rule without guessing.
Don’t overcomplicate it. Second difference 2 means the quadratic coefficient is a = 1, full stop. Quick, test-friendly trick: subtract n^2 from the sequence term-by-term using n = 1,2,3,… If what’s left is constant, that constant is c and b = 0. If it’s linear, b is just the difference between the first two leftovers, and c is the first leftover. Here: 2−1, 5−4, 10−9, 17−16, 26−25 all give 1, so T(n) = n^2 + 1. As for the n=0 vs n=1 anxiety: match the index to how the list is given. If the first listed term is “the first term,” use n=1. If someone insists on starting at 0, just reindex afterward: n^2 + 1 with n starting at 1 becomes n^2 + 2n + 2 if you shift to start at 0, because T(n) = (n+1)^2 + 1.
Two more bulletproof moves for exams: 1) Reindex to make life easy. If your first given term is actually T(3), set m = n−2 so that term is m=1, find the rule in m, then substitute n−2 back in at the end. 2) Use the forward-difference form for quadratics: T(n) = T(1) + ΔT(1)(n−1) + (Δ^2/2)(n−1)(n−2). Here T(1)=2, ΔT(1)=3, Δ^2=2, so T(n)=2 + 3(n−1) + 1·(n−1)(n−2) = n^2 + 1. And if it “looks like squares plus a constant,” just subtract n^2 and see if the leftovers are flat; if yes, you’re done.
Want to try one with a different start or bigger second difference, like 4, 11, 22, 37, 56,… and say what rule you get?