I never was taught about arithmetic, geometric, or any other type of progression or pattern. Yet, the SATs love these! Well, I took my lumps ages ago and skipped these problems. But, now I ask on behalf of my homeschooled daughter. (Yes, a rare topic I cannot teach.) What is/are the math tricks to crack the patterns? Ok, the arithmetic and geometric progressions are not so bad. But, she’s working from a Barron’s guide that doesn’t always explain things very well. They throw her into a combo of arithmetic-geometric pattern that difficult to crack.
The problem is 2, 4, 10, 28, 82… The only way I was taught (for the very little time we spent on this) would be to find the deltas between the elements given. This gave me 2,6,18,54… = 3^n*2 for n=0 to… Thus, what I actually found was the pattern of the deltas, not the original pattern! So, how would you find the pattern?
I don’t really know how you to empirically find it. For me, I just sort of see/guess whether its a polynomial, an exponential, or factorial, then try combinations until I find it.
ETA: In this case, I guess the first step would be figuring out any algebraic shift. So you would have to notice that 28 is 1 more than 27 and 82 is 1 more than 81 and so forth. If you reduce all the numbers by one, the sequence (series?) becomes 1, 3, 9, 27, 81… which is pretty simple to figure out.
Polynomials are easy. Keep taking deltas of deltas until you get a constant and the number of iterations is the degree of the polynomial.
0, 9, 30, 69, 132, 225
1st deltas: 9, 21, 39, 63, 93
2nd deltas: 12, 18, 24, 30
3rd deltas: 6, 6, 6, 6
3 iterations means a third power and because the delta is 6, the coefficient of the cubic term is 6/(3!*) = 1
subtract 1 x n[sup]3[/sup] from each term in the original sequence
0-1, 9-8, 30-27, 69-64, 132-125, 225-216
-1, 1, 3, 5, 7, 9
lather rinse repeat
n[sup]3[/sup] + 2n - 3
*because Calculus. If the term is of degree d, divide the delta by d! to get the coefficient.
For the exponential, taking the deltas once* eliminates the constant term. We can worry about that later. You correctly identified that it is a geometric sequence so it is exponential. Let’s assume a basic exponential function a x 3[sup]u[/sup] + v (otherwise it would be nearly intractable). That mean there should be a constant that you can add or subtract to each term to get a constant time the powers of 3. In this case, subtract 1 from each term and you get 1, 3, 9, 27, 81 or 3^0, 3^1, 3^2, 3^3, 3^4. That is the pattern of 3[sup]n-1[/sup]. Add 1 back to each term and there is the answer The Joker and the Thief gave.
Example that shows each step (your example uses 1x the powers of 3s so it really is missing a step): 194, 770, 3074, 12290
deltas: 576, 2304, 9214
each delta is 4x the previous one so I know it is a x 4[sup]u[/sup]
what number do I add or subtract to the terms of the sequence so each one is exactly 4x the previous one? In this case subtract 2 to get 192, 768, 3072, 12288
The highst power of 4 that divides 192 is 4[sup]3[/sup]. Since 192 = 3 x 4[sup]3[/sup] we tie everything together to say An = 3 x 4[sup]n+2[/sup] + 2
ETA: For The Joker and the Thief, it is at this point that determines if it is exponential viz, the first iterations of deltas is itself a geometric sequence.
The way I do these problems is more quick and dirty. When I see the deltas as
2, 6, 18, 54 I see an immediate pattern. 23 =6. 63 = 18. 183 = 54. 543 = 162. 162 + the last number on our list, 82 = 244, which arrives at the same answer as 1+3^(6-1). If I have a few seconds to spare, then I can work out the 1+3^*(n-1) equation, but I tend to race through these things and see if an answer fits the pattern I’ve already found. If I saw 244 on the list, I’d fill it in and go on to the next one.
Putting 2, 4, 10, 28, 82 into the search box at OEIS yields seven possible series, with the next entry being 244, 244, 248, 255, 260, 276, and 264.
Therefore, what matters is context. For a high school student the probability is high that the formula being sought is 3^n + 1 rather than the Catalan numbers, especially if the section is on exponential sequences.
I was halfway joking about the Catalan thing and halfway pointing out that there’s always more than one way to look at these what is the next number in the sequence problems.
Right. You and Lance Turbo are making the same point. There’s no general, all-purpose mechanical rule for “Here are some numbers. What comes next?”. The sequence could extend any which way, any which way at all! It’s up to you to decide which way seems like the nicest pattern to you, a semi-subjective assessment.
(For just this reason, and also in keeping with my memory of it, I highly doubt the SAT has questions of precisely this “Here are some number. What comes next?” form. I associate that more with “IQ tests”. The SAT might conceivably ask you “What comes next in this arithmetic sequence?” or “What comes next in this geometric sequence?”, but by naming the type of sequence, they are also essentially telling you its basic rule, instead of requesting that you read their mind.)
I believe the proof that the next number can be any real number is:
Assume m terms such that f(n) can describe the terms. Let g(n) = k(n-1)(n-2)(n-3)…(n-m) + f(n).
g(n) can also describe the sequence.
What’s the next number in this sequence? 1, 3, 5, 7
Answer 42
The function is (11/40)(n-1)(n-2)(n-3)(n-4) + (2n-1)
To expand on the above, any sequence of m terms can be described by a polynomial of degree m+1 and through a system of equations you can actually find the coefficients of the polynomial. Call this polynomial f(n). Let j = the value of f(m+1) and x be the value you want to generate. Then the k in g(n) from above is [x-f(m+1)]/[(m+1)!]
<classroom lawyer>That’s not what the question said AND you changed my formula! Where is it implicit that we use a polymorphic formula?</classroom lawyer>
The proof that the next number can be anything is that you can go ahead and write down anything next. Nothing’s stopping you.
It is also interesting to note, as you do, that even restricting to polynomials, the next number could be anything, if we do not bound the degree of the polynomial sufficiently. But regardless, typically these “What number comes next?” riddles never explicitly say anything about polynomials in the first place. [And, indeed, if the sequence which had appeared before us originally was “1, 10, 100, 1000”, few of us would feel the need to stick to polynomials in extrapolating it.]
Place two dots on the perimeter of a circular pie, connect the dots and divide the pie into 2 pieces. Connect 3 dots and get 4 pieces; 4 dots, 8 pieces; and so on.
