If you don’t know how to find the n^{th} 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 n^{th} term in quadratic sequences. For those who don’t, here’s a quick run-through:

To find n in a sequence with an n^{th} term of the form an^{2} + 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 n^{2}, or the value of a in an^{2} + bn + c.

4. Now generate the values of an^{2}, 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. (an^{2} + bn + c – an^{2} = bn + c)

This may not have been entirely clear. If you wish, you may read the following example:

Find the n^{th} 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:

3n^{2}: 3, 12, 27, 48

5.Original sequence – (4.) = (13-3), (26-12), (45-27), (70-48)

= 10, 14, 18, 22

6. Find n^{th} term of (5.): 4n+6

Put this back with the quadratic part we found earlier:

3n^{2}+4n+6

You may, of course, wonder how one would go about finding the n^{th} term in a cubic sequence (of the form an^{3} + bn^{2} + 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 n^{3} but we have to divide by something first. What do you think it is?

With a linear sequence we find the 1^{st} difference and don’t divide by anything (or divide by 1). With a quadratic sequence we find the 2^{nd} difference and divide by 2. With a cubic we find the 3^{rd} 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: an^{k} + bn^{k-1} + … + yn + z:

Find the k^{th} difference between the terms and divide by k!.

Stated as a formula:

Let’s try an example to clarify things.

Find the n^{th} term of the following sequence:

8, 88, 562, 2228, 6604, 16168, 34598

1. Find the differences:

d_{0} |
8 88 562 2228 6604 16168 34598 |

d_{1} |
80 474 1666 4376 9564 18430 |

d_{2} |
394 1192 2710 5188 8866 |

d_{3} |
798 1518 2478 3678 |

d_{4} |
720 960 1200 |

d_{5} |
240 240 |

2. As the difference has become a constant at the 5^{th} difference (d_{5}), this will relate to the coefficient of n^{5}. So the value of the a in an^{5} is d_{5} ÷ 5!. d_{5} ÷ 5! = 240 ÷ 120 = 2

Thus, the coefficient of n^{5} is 2 and our first term is 2n^{5}.

3. Subtract the 2n^{5} 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:

d_{0} |
6 24 76 180 354 616 984 |

d_{1} |
18 52 104 174 262 368 |

d_{2} |
34 52 70 88 106 |

d_{3} |
18 18 18 18 |

5. As the difference has become a constant at the 3^{rd} difference (d_{3}), this will relate to the coefficient of n^{3}. The value of a from an^{3} is d_{3} ÷ 3!. d_{3} ÷ 3! = 18 ÷ 6 = 3

Thus, the coefficient of n^{3} is 3 and our second term is 3n^{3}.

6. Generate the sequence of 3n^{3} 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 n^{th} term for that particular sequence is 2n^{5} + 3n^{3} – n^{2} + 4

Next, we can look at the proof:

The n^{th} term of a simple quadratic sequence would contain an^{2}, 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}, an^{2}, a(n+1)^{2}

Firstly, expand the binomials:

a(n^{2}-2n+1), an^{2}, a(n^{2}+2n+1)

Then find the differences:

d_{1} |
a(2n-1), a(2n+1) |

d_{2} |
2a |

Thus dividing by two reveals the coefficient, a.

A similar proof can be used for cubic sequences:

Terms: a(n-1)^{3}, an^{3}, a(n+1)^{3}, a(n+2)^{3}

Expand: a(n^{3}-3n^{2}+3n-1), an^{3}, a(n^{3}+3n^{2}+3n+1), a(n^{3}+6n^{2}+12n+8)

d_{1}: a(3n^{2}-3n+1), a(3n_{2}+3n+1), a(3n_{2}+9n+7)

d_{2}: 6an, 6an+6a

d_{3}: 6a

This, again, proves that the third difference of the terms of a cubic sequence will be six times the coefficient of n^{3}.

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 n^{x} 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.