Solutions to Challenge #3 follow. If you absolutely must read the solutions without attempting the challenge, I certainly can’t stop you. But Santa Claus knows.
- To get each term from the previous term, multiply by two and add one. If we use the notation to denote the nth term, we can express the solution like this:
In other words, the first term is equal to 1 (), and to get the nth term (), multiplty the previous, i.e. (n-1)st, term by two and add one (). So the next three terms are 63, 127, 255.
- The second sequence can be described in a similar fashion:
I will let you work out a corresponding description in English; the next three terms are 365, 1094, 3281.
- The easiest way to describe the third sequence is that the nth term is equal to n times (n + 1), i.e. the sequence can be written as . Written explicitly, . Hence the next three terms are 42, 56, 72. But there are other equivalent ways to describe this sequence. In fact, this is equally true of the first two sequences as well. Further challenge: can you come up with other ways to characterize these three sequences? How many different characterizations can you come up with?
- To get each term from the previous term, add up the digits, and multiply by two. For example, . Therefore the next three terms are 8, 16, 14. Notice that if you keep going further, the sequence enters a loop: . Further challenge: will this always happen? In other words, picking any starting value you want, and applying the same rule as this sequence, will the sequence always enter a loop? How many different loops are there?
hey mr yorgey, whats up?
this stuff is hard!
Hey Jeremy! By “hard” do you mean it’s difficult but you understand it, or do you mean you don’t understand it? If it’s the latter of course I’d be happy to try to explain better whatever it is you don’t get…