I’m revising for a test and I’m stuck (again) on finding the nth term for a quadratic sequence. My brain keeps doing cartwheels over the a, b, c bit.
Example I’m practicing: 1, 6, 15, 28, 45, …
– First differences: 5, 9, 13, 17
– Second differences: 4, 4, 4
So I think that means a should be 2 (half the second difference), right?
Then I try to get b. I used the idea that the first difference at the start is 2a + b (but I might be misremembering this). So I did 5 = 2*2 + b, which gave b = 1. Then I used the first term to get c: 1 = a + b + c = 2 + 1 + c, so c = -2. That would make the formula 2n^2 + n – 2.
But when I plug in n = 3, I get 2*9 + 3 – 2 = 19, and the actual third term is 15. So… something is off. Am I supposed to use 3a + b for the first difference instead of 2a + b? Or is my mistake actually that I’m starting at n = 1 when I should be starting at n = 0 (or vice versa)?
I also tried the subtract-the-quadratic trick: subtract 2n^2 from each term. That gave me -1, -2, -3, -4, … which looks like “minus something linear,” but I can’t tell if I should read that as -n or -(n – 1) or what. My head’s doing the off-by-one cha-cha.
Could someone explain which expression the first difference should equal after I find a, and how to line up the n’s properly so I stop being 1 off? Any help appreciated!
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3 Responses
You’re right that a = 2 since the second difference is 4; for T_n = an^2 + bn + c we have T_{n+1} − T_n = 2an + a + b, so the first gap is 3a + b = 5 giving b = −1, and using T_1 = 1 gives c = 0, hence T_n = 2n^2 − n.
Would you like a quick derivation of T_{n+1} − T_n = 2an + a + b (and why starting at n = 1 makes the first gap 3a + b)?
You’re so close! Your a = 2 is spot on because the second difference is 4, and that “half the second difference” trick is exactly right. The hiccup is the first-difference formula: for T(n) = an^2 + bn + c (with the first term at n = 1), the nth first difference is D(n) = T(n+1) − T(n) = a(2n + 1) + b. That means the very first difference (the 5) is D(1) = 3a + b, not 2a + b. So plug in a = 2: 3a + b = 5 gives 6 + b = 5, so b = −1. Then use the first term: 1 = a + b + c = 2 − 1 + c, so c = 0. That gives T(n) = 2n^2 − n, which hits 1, 6, 15, 28, 45 perfectly (check n = 3: 18 − 3 = 15). Your subtract-2n^2 method was also whispering the same thing: the leftovers were −1, −2, −3, … which is just −n when you start at n = 1. Think of it like stairs where each step gets 4 higher than the last; the “extra +a” in 2an + a + b is that built-in first riser people often forget. If you ever start indexing at n = 0, then the very first difference would be a + b instead-so the off-by-one is all about where you start counting.
Your cartwheeling brain is onto something: the constant second difference 4 means 2a = 4, so a = 2. For a quadratic tn = an^2 + bn + c (with n starting at 1), the first differences form an arithmetic sequence: Δt_n = t_{n+1} − t_n = a(2n + 1) + b. That means the very first difference (between t1 and t2) is 3a + b. So here 5 = 3·2 + b, giving b = −1. Now use the first term: 1 = a + b + c = 2 − 1 + c, so c = 0. Ta‑da: tn = 2n^2 − n. Quick check: n = 3 gives 18 − 3 = 15, which matches. The “2a + b” ghost you remembered isn’t the first difference at n = 1; the constant second difference is 2a, and the first first-difference is 3a + b when you start counting at n = 1.
Your subtract-the-quadratic trick was dancing in the right shoes, too. Subtract 2n^2 from each term: −1, −2, −3, −4, … That’s exactly −n when n starts at 1, so the leftover linear part is −n + 0, giving b = −1 and c = 0 again. The off-by-one cha-cha usually happens when mixing “start at n = 0” with “start at n = 1”; pick one and stay consistent, and the numbers behave. Want to try another sequence and see if you can spot 3a + b in the first difference and 2a in the second difference straight away?