If you don’t know how to find the nth term in a standard linear sequence then I’m ashamed of you, leave now. If you do, you may also know how to find the nth term in quadratic sequences. For those who don’t, here’s a quick run-through:
To find n in a sequence with an nth term of the form an2 + bn + c:
1. Find the difference between each term that you’ve been given.
2. Find the difference between each difference of each term that you’ve been given.
3. This should now be a constant. (If it’s 0, the sequence is linear. If it isn’t a constant, the sequence isn’t quadratic.) Divide it by 2. This number is the coefficent of n2, or the value of a in an2 + bn + c.
4. Now generate the values of an2, now that you know the value of a.
5. Subtract your original values from the ones in this new sequence.
6. This will leave the linear part of the sequence which can be solved easily. (an2 + bn + c – an2 = bn + c)
This may not have been entirely clear. If you wish, you may read the following example:
Find the nth term of the sequence:
13, 26, 45, 70
1. First difference: (26-13), (45-26), (70-45)
= 13, 19, 25
2. Second difference: (19-13), (25-19)
= 6, 6
3. Divide this constant by 2: 6/2=3
4. New sequence:
3n2: 3, 12, 27, 48
5.Original sequence – (4.) = (13-3), (26-12), (45-27), (70-48)
= 10, 14, 18, 22
6. Find nth term of (5.): 4n+6
Put this back with the quadratic part we found earlier:
You may, of course, wonder how one would go about finding the nth term in a cubic sequence (of the form an3 + bn2 + cn + d). You will also, probably, guessed a few of the first steps. Take the first difference, the second difference and then the third difference. This should give us the coefficent of n3 but we have to divide by something first. What do you think it is?
With a linear sequence we find the 1st difference and don’t divide by anything (or divide by 1). With a quadratic sequence we find the 2nd difference and divide by 2. With a cubic we find the 3rd difference and divide by what?
Let’s look at the sequence so far: 1, 2, 6
Got it yet?
If not, here’s the next term: 24
You should be able to get it by now.
Don’t worry if you can’t, here’s the next: 120
1, 2, 6, 24, 120, …, n! (n factorial)
1×2=2, 1×2×3=6, 1×2×3×4=24, 1×2×3×4×5=120
To generalise: To find the coefficient of a sequence of the form: ank + bnk-1 + … + yn + z:
Find the kth difference between the terms and divide by k!.
Stated as a formula:
Let’s try an example to clarify things.
Find the nth term of the following sequence:
8, 88, 562, 2228, 6604, 16168, 34598
1. Find the differences:
|d0||8 88 562 2228 6604 16168 34598|
|d1||80 474 1666 4376 9564 18430|
|d2||394 1192 2710 5188 8866|
|d3||798 1518 2478 3678|
|d4||720 960 1200|
2. As the difference has become a constant at the 5th difference (d5), this will relate to the coefficient of n5. So the value of the a in an5 is d5 ÷ 5!. d5 ÷ 5! = 240 ÷ 120 = 2
Thus, the coefficient of n5 is 2 and our first term is 2n5.
3. Subtract the 2n5 from the original sequence:
8, 88, 562, 2228, 6604, 16168, 34598
- 2, 64, 486, 2048, 6250, 15552, 33614
= 6, 24, 76, 180, 354, 616, 984
4. Find the differences of this new sequence:
|d0||6 24 76 180 354 616 984|
|d1||18 52 104 174 262 368|
|d2||34 52 70 88 106|
|d3||18 18 18 18|
5. As the difference has become a constant at the 3rd difference (d3), this will relate to the coefficient of n3. The value of a from an3 is d3 ÷ 3!. d3 ÷ 3! = 18 ÷ 6 = 3
Thus, the coefficient of n3 is 3 and our second term is 3n3.
6. Generate the sequence of 3n3 and subtract it from the sequence generated in (3.):
6, 24, 76, 180, 354, 616, 984
- 3, 24, 81, 192, 375, 648, 1029
= 3, 0, -5, -12, -21, -32, -45
7. Continue solving this in the same manner until there are no other terms left.
The nth term for that particular sequence is 2n5 + 3n3 – n2 + 4
Next, we can look at the proof:
The nth term of a simple quadratic sequence would contain an2, with coefficient a. Excluding all other terms which would cancel each other out by the second difference, the terms would be:
a(n-1)2, an2, a(n+1)2
Firstly, expand the binomials:
a(n2-2n+1), an2, a(n2+2n+1)
Then find the differences:
Thus dividing by two reveals the coefficient, a.
A similar proof can be used for cubic sequences:
Terms: a(n-1)3, an3, a(n+1)3, a(n+2)3
Expand: a(n3-3n2+3n-1), an3, a(n3+3n2+3n+1), a(n3+6n2+12n+8)
d1: a(3n2-3n+1), a(3n2+3n+1), a(3n2+9n+7)
d2: 6an, 6an+6a
This, again, proves that the third difference of the terms of a cubic sequence will be six times the coefficient of n3.
The same concept will provide proofs for quartic, quintic, sextic, etc… sequences. You cannot, however, continuously rely on individual proofs for each type of sequence, you must generalise these proofs which supply evidence ad infinitum. This, I have not done. I have been somewhat busy and thus I leave this possibility open for anyone to look into further. Please inform me if you do.
- Ruskin Harding, Realms of Obscurity
P.S.: To find the coefficient of nx you must be given at least x+1 terms. If you already know the value of x, you can solve it with just x terms.