I’m revising my algebra fundamentals and trying to get a clearer feel for quadratic sequences. I keep hearing: constant second differences mean it’s quadratic, and that constant equals 2a if the nth term is an^2 + bn + c. I sort of believe it, but I don’t see why it’s specifically 2a. Where does that 2 come from in the differences?
Example I’m playing with: 3, 8, 15, 24, 35. First differences are 5, 7, 9, 11. Second differences are 2, 2, 2. So I think a should be 1. Then I try to find b and c and my notes turn into spaghetti. I tried subtracting n^2 from each term (for n = 1, 2, 3, 4, 5) and I got 2, 4, 6, 8, 10, which looks like 2n – is that a legit move or just me pattern-hunting? If it is legit, how do I turn that into a clean expression for the whole sequence? If it’s not, what’s a better, more reliable path?
Also, tiny side confusion: does it matter if I index from n = 0 vs n = 1? I think I’m mixing those up and getting different constants.
Could someone explain a simple, repeatable way to (1) justify the ‘second difference = 2a’ idea, and (2) systematically get a, b, c for a sequence like the one above? I’m really trying to strengthen the basics so I don’t have to guess on test problems.
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3 Responses
Great question. Two things to tackle:
– why the second difference is exactly 2a for an^2 + bn + c, and
– a clean, repeatable way to extract a, b, c (and how indexing affects them).
I’ll go step by step.
1) Why the second difference is 2a
Let Δ denote the first-difference operator: Δf(n) = f(n+1) − f(n). The second difference is Δ^2 f(n) = Δ(Δf(n)) = f(n+2) − 2f(n+1) + f(n).
Now compute this for f(n) = an^2 + bn + c:
– f(n+1) = a(n+1)^2 + b(n+1) + c = a(n^2 + 2n + 1) + bn + b + c
– f(n+2) = a(n+2)^2 + b(n+2) + c = a(n^2 + 4n + 4) + bn + 2b + c
Plugging into Δ^2:
f(n+2) − 2f(n+1) + f(n)
= [a n^2 + 4a n + 4a + b n + 2b + c]
− 2[a n^2 + 2a n + a + b n + b + c]
+ [a n^2 + b n + c]
Group like terms. All n^2 and n terms cancel:
– an^2 − 2an^2 + an^2 = 0
– (4an) − 2(2an) = 0
– (bn) − 2(bn) + (bn) = 0
Constants: 4a + 2b + c − 2(a + b + c) + c = 4a + 2b + c − 2a − 2b − 2c + c = 2a.
So Δ^2(an^2 + bn + c) = 2a, independent of n. That “2” is the discrete analogue of taking two derivatives: the jump from (n+1)^2 to n^2 contains a 2n term, and doing it twice strips off the n, leaving 2a.
A couple of quick checkpoints:
– Δ(n) = 1, so Δ^2(n) = 0.
– Δ(constant) = 0, so Δ^2(constant) = 0.
– Linearity: Δ^2(an^2 + bn + c) = a Δ^2(n^2) + b Δ^2(n) + c Δ^2(1) = a·2 + b·0 + c·0 = 2a.
2) A systematic recipe to find a, b, c
You can do this in several equivalent ways. Here are two that are fast and reliable.
Method A: Differences + “strip off” the quadratic part
– Step 1: Compute the constant second difference d. Then a = d/2.
– Step 2: Form the residual r_n = s_n − a n^2. This must be linear: r_n = bn + c.
– Step 3: The first difference of r_n is constant and equals b (since Δ(bn + c) = b). So compute any r_{n+1} − r_n to get b.
– Step 4: Solve for c using any term, e.g. c = r_1 − b·1 (if you’re indexing from n = 1).
Method B: Three quick formulas (when indexing from n = 1)
Let s_1, s_2, s_3 be the first three terms, and let d be the constant second difference.
– a = d/2.
– b = (s_2 − s_1) − 3a. Reason: s_2 − s_1 = (4a + 2b + c) − (a + b + c) = 3a + b.
– c = s_1 − a − b. Reason: s_1 = a + b + c.
Both methods give the same result.
Worked example: 3, 8, 15, 24, 35 (indexed from n = 1)
First differences: 5, 7, 9, 11.
Second differences: 2, 2, 2. So d = 2 and a = d/2 = 1.
Using Method A (strip off the quadratic):
– Residual r_n = s_n − n^2:
n = 1: 3 − 1 = 2
n = 2: 8 − 4 = 4
n = 3: 15 − 9 = 6
n = 4: 24 − 16 = 8
n = 5: 35 − 25 = 10
So r_n is 2, 4, 6, 8, 10, which is an arithmetic sequence with common difference 2.
– Therefore b = Δr_n = 2.
– Then c = r_1 − b·1 = 2 − 2 = 0.
Conclusion: s_n = n^2 + 2n + 0 = n^2 + 2n. Check: for n = 5, 25 + 10 = 35. Works.
Your “subtract n^2 and spot 2n” move was legitimate because:
– you first determined a = 1 from the second differences,
– subtracting a n^2 must leave a linear sequence bn + c,
– the data showed that linear sequence is exactly 2n (so b = 2, c = 0).
What would not be legitimate is subtracting some guessed quadratic before confirming a from the second differences.
Indexing: starting at n = 1 vs n = 0
– The second difference and the value of a do not depend on where you start indexing. So a is stable.
– b and c do depend on the starting index. Shifting n to n+1 changes the linear and constant terms:
If t_k = s_{k+1}, then t_k = a(k+1)^2 + b(k+1) + c = a k^2 + (2a + b)k + (a + b + c).
So a stays the same, while b and c shift to b’ = 2a + b and c’ = a + b + c.
– Practical advice: pick an index convention and stick with it throughout the calculation. If you start at n = 1, use the “Method B” formulas above. If you start at n = 0, the quick relations adapt to:
a = d/2,
b = (s_1 − s_0) − a, because s_1 − s_0 = a + b,
c = s_0, because s_0 = c.
A second quick example (indexed from n = 0)
Suppose a sequence starts 4, 3, 6, 13, 24, … with n = 0, 1, 2, 3, 4.
– First differences: −1, 3, 7, 11.
– Second differences: 4, 4, 4. So d = 4, hence a = 2.
– Using the n = 0 formulas:
b = (s_1 − s_0) − a = (3 − 4) − 2 = −1 − 2 = −3.
c = s_0 = 4.
– Therefore s_n = 2n^2 − 3n + 4. Check n = 3: 2·9 − 9 + 4 = 18 − 9 + 4 = 13, matches.
Summary cheat sheet
– Constant second difference d ⇒ quadratic. The constant equals 2a because Δ^2(an^2 + bn + c) = 2a.
– From differences (n = 1 indexing):
a = d/2,
b = (s_2 − s_1) − 3a,
c = s_1 − a − b.
– Or: subtract a n^2 to get a linear residual, read off b from its first difference, then solve c from any term.
– Index shifts don’t change a or the second difference, but they do change b and c. Decide on n = 0 or n = 1 at the start and stay consistent.
With these steps, you won’t need to guess on test problems-you can compute a, b, c directly and check quickly.
Think of a quadratic sequence like a car’s position each second when it’s speeding up at a steady rate: the “first differences” are like velocities, and the “second differences” are like the constant acceleration-so if the nth term is an^2 + bn + c, the second differences come from how (n+1)^2 − n^2 expands to 2n + 1, and doing that difference one more time wipes out the n, leaving exactly 2a every time. For your sequence 3, 8, 15, 24, 35 the second difference is 2, so 2a = 2 and a = 1. A super-clean way to find b and c is exactly what you tried: subtract n^2 from each term (for n = 1, 2, 3, …), because T_n − n^2 = bn + c must be linear; you got 2, 4, 6, 8, 10, which is 2n, so bn + c = 2n gives b = 2 and c = 0. Quick check: T_n = n^2 + 2n gives 3, 8, 15, 24, 35 for n = 1 to 5-bullseye. Alternatively, you can solve small equations: plug n = 1, 2, 3 into an^2 + bn + c = 3, 8, 15, use a = 1, then solve b + c = 2, 2b + c = 6 to get b = 2, c = 0. Your “subtract n^2” move is totally legit and very reliable because it peels the quadratic down to a line, whose first difference is constant. About indexing: it doesn’t change a, and usually only shifts c while b stays put-so starting at n = 0 vs n = 1 can tweak the constants but not the overall shape. In short, the “2” appears because of the (n+1)^2 = n^2 + 2n + 1 expansion, and the repeatable recipe is: read off a from the constant second difference via a = (second difference)/2, then remove an^2 to expose a line and fit b and c from that.
Because (n+1)^2 − n^2 = 2n + 1, the first difference of an^2 + bn + c is 2an + (a + b), and differencing that again cancels the n to give 2a; starting at n = 0 or n = 1 only changes b and c, not a-just be consistent. For 3, 8, 15, 24, 35 the second difference is 2, so a = 1; subtract n^2 (with n = 1, 2, 3, …) to get 2, 4, 6, 8, 10 = 2n, hence b = 2 and c = 0, so the nth term is n^2 + 2n-and I still remember the day I realized this “subtract the quadratic, read off the linear” trick is like a discrete derivative and it felt like magic!