
Finding
the Next Number in a Sequence: Sections: Common differences, Recursions, General examples, Nonmath "sequences"
I could try to cast about for a formula for this sequence, but I think I'll go right to finding differences: Since the third differences are all the same (in this case, "6"), the generating polynomial for this sequence is a cubic, or third degree, polynomial, so it is of the form: an^{3} + bn^{2} + cn + d I'll evaluate this at some of the values of n: a(1)^{3} + b(1)^{2} + c(1) + d = a + b + c + d = 0 This gives me four equations in four unknowns. I'll use matrices again to solve: Then a = 1, b = –13, c = 44, and d = –32, so the formula is: 1n^{3} + (–13)n^{2} + 44n + (–32) = n^{3} – 13n^{2} + 44n – 32 They gave me the first six terms of the sequence, so I need to give the seventh term, which is:
= 343 – 637 + 308 – 32 = –18
You might notice right away that every term is gotten by multiplying the previous term by 2. That is, the formula in this case is not a polynomial, but an exponential: a_{n} = 2^{n} In mathspeak, the nth term a_{n} ("aysubenn") is 2 to the power n. But what if you didn't happen to notice this? What would happen if you tried to find the differences? I will try to find the common difference: As you can see, I am not arriving at a common difference. However, do you see something interesting in the diagonal rows? In each toplefttobottomright diagonal, the values are the same! For sequences where the rule is "multiply by some fixed amount", the diagonals are always multiplied by one less than this fixed amount. Since the rule for this sequence is "multiply by 2", the entries on the diagonals are multiplied by 1, which is why they are unchanged going down the diagonals. In either case, I came up with the rule "multiply the last term by 2 to get the next term", so the next term here is:
In this case I am almost multiplying by 3, but "multiply by 3" doesn't account for the minus sign. Notice, though, that the minus sign alternates; only every other term is negative. This means that we are multiplying by a negative three, so the next term is:
By the way, the differences in this case look like this: Going down each diagonal (highlighted in color, above), you can find the multiplier for each row by taking the lower number and dividing by the higher number. For instance, 12 ÷ –3 = –4 and 144 ÷ –36 = –4. Since the multiplier for the table of differences is one less than the multiplier for the sequence, then the sequence multiplier is one more than the table multiplier, or –4 + 1 = –3, which we already knew.
These numbers aren't growing really quickly, so this doesn't appear to be a "multiply by some fixed value" kind of sequence. Supposing I didn't notice right away that the rule is "add two to the previous value", I can still find the answer from the table of differences: Since the first differences are the same, this means that the rule is a linear polynomial, something of the form y = an + b. I will plug in the first couple of values from the sequence, and solve for the coefficients of the polynomial: 5 = 1a + b This system solves as: So the formula is y = 2n + 3. Since they gave me the first five terms, the next term will be the sixth term: Copyright © Elizabeth Stapel 20022011 All Rights Reserved
This doesn't appear to be a "multiply by a fixed value" kind of sequence, and there doesn't appear to be a fixed value that is added each time, either. (For instance, in the last sequence, we always added 2 to get the next term.) Let's see what the differences give me: Since the second differences are the same, the formula for this sequence is a quadratic, y = an^{2} + bn + c. I'll plug in the first three terms for y and solve for the values of a, b, and c: 1a + 1b + c = 1 Then the formula for this sequence is ( ^{1}/_{2} )n^{2} + ( ^{3}/_{2} )n – 1. Since they gave me the first six terms, the next term will be the seventh: ( ^{1}/_{2} )(7)^{2} + ( ^{3}/_{2} )(7) – 1 = ^{49}/_{2} + ^{21}/_{2} – 1 = 34 << Previous Top  1  2  3  4  5  6  7  Return to Index Next >>


MATHHELP LESSONS
This lesson may be printed out for your personal use.

Copyright © 20022014 Elizabeth Stapel  About  Terms of Use  Linking  Site Licensing 




