More Worked Examples (complete with thought process)

These fractions are not equal distances apart, so adding the same thing to each term is not the rule. The fractions aren't multiplied by the same value, and they're not powers.

Well, I mean, the whole fractions aren't powers. The denominators look like powers of 2, but the numerators look like a list of odd numbers. Hey....

Sometimes, especially when dealing with a sequence of fractions, you may discover that they've really given you two sequences in one. In this case, looking at the numerators and denominators separately, I find this:

The numerators appear to be counting by twos; the denominators appear to be doubling. So the missing term is:

Aw, geez; when they give me really long lists like this, that means that I *need* the long list to have any hope of figuring out the rule. Which means that, whatever the rule turns out to be, it's probably not going to be easy to figure out.

The third term is the cube of the second term, but no other pair shows an exponential relationship. I'm not seeing squares or cubes, other than the 4 being the square of 2, and the 4 is the sixth term in the list, which doesn't seem to fit anything.

It doesn't help that the values go up and down repeatedly. I could try finding the differences, but I think I'd end up with a degree-12 polynomial. If I can't find anything else, I'll do the polynomial, but — yeesh! I hope I can find something easier. Like, I see 2, 3, 4, 5, and 6, listed in order, every other term, but... wait....

What are the other terms, taking out the 2, 3, 4, 5, and 6? I get:

Ah! These numbers are three apart. This sequence is really two in one. If I take the sequence terms apart, listing alternate terms in their own sequences, I get:

In other words, the odd-numbered terms (the first, third, fifth, etc) appear to be of the form "add 3 to the previous term", and the even-numbered terms (the second, fourth, sixth, etc) appear to be of the form "add 1 to the previous term". Then the next two terms are 17 + 3 = 20 and 6 + 1 = 7.

If I try common differences on this sequence, I find that I'd need a degree-7 polynomial. I'd like to avoid that. Whenever a sequence is not obviously generated by a polynomial, it is often (though not always) generated by a recursion. That is, I should look for some pattern that relates the later terms in this sequence to the earlier ones. I'll try adding terms:

Hmm... The last sum is not 7, so that doesn't work. And multiplication clearly doesn't work, since 2×3 = 6, not 5. However, I notice that:

I would guess that the formula is meant to be:

There is no guarantee that this is the "right" answer — in the sense of being the answer that the author had in mind — but this formula does "work", so it should be acceptable. (It's weird, though, that the first three terms are essentially distractors.) Following my rule, the next number is 6 + 13, which equals:

I did common differences with the first five terms, and got a fourth-power polynomial. But I checked, and it didn't give me the seventh term, so this doesn't appear to be a polynomial sequence. The terms are growing quickly, but not nearly fast enough to be factorials. But these don't appear to be multiples of anything, either.

It looks like, the further I go, the closer the pairs get to being multiples of 3. But they're not quite multiples of 3. Can I express them as multiples, but tweaked a bit?

This doesn't appear to be working. What if the 2 is a seed value, and this is a recursive? Can I do better with multiples of 3 if I multiply by the previous term?

If I add the index, I think I've found the pattern:

It looks like the rule may be "multiply the previous value by 3, and then add the value of the index". If this recursive rule fits, then my next two terms will be:

And this fits! It gives me a value for the missing term, while matching the last term! Whoo-hoo!

So my pattern is:

...and my answer for the missing term is:

For this one, I noticed that the first and third term appear to be related, as do the second and fourth:

But trying to pair the third and fifth term this way doesn't work, because 2×4 equals 8, not 6. Is there another pattern for the next pair of terms? Well, 3×2 gives me the second 6. What if I multiply the second term by the same thing?

So maybe the 2 and the 3 are seed values, and you multiply these values by bigger and bigger numbers, in pairs. If so, then the next multiple would be 4. Does this check?

This looks like it fits. And the next number is then 4×3 = 12. I'm going to use this for my answer: