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Numerade Educator



Problem 68 Hard Difficulty

Find the first 40 terms of the sequence defined by
$ a_{n + 1} =\left\{
\frac{1}{2} a_n & \text{if } a_n \text{ is an even number} \\
3a_n + 1 & \text{if } a_n \text{ is an odd number } \end{array} \right. $
and $ a_1 = 11. $ Do the same if $ a_1 = 25. $ Make a conjecture about this type of sequence.


The terms of this sequence will eventually be reduced to having a greatest odd factor of $1,$ regardless of the positive
starting value. $\$ \$$


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Video Transcript

find the first forty terms of the sequence given by the following formula. In case one let's look at a one equals eleven a two. This will be three a one plus one, since a one isn't even a nod number. Excuse me when it's a remember use. This Formula eleven was odd. So we use this formula here, and that's how we got a two. So it's kind of the reasoning. They're the heroes indicate the reasoning to find the following values of the sequence. So this is just three times eleven plus one thirty four now a three since thirty four's An even number. We divided by two thirty four even use this formula. And then that brings us to why we use this formula over here. And I want a spell that out every time. But if it's an even or odd number that tells me which formula to use because it's peaceful, eyes function a four that's three times seventeen plus one. So fifty two a five Well, we have an even number. So we'll divide that by two. Have another even so, also, divide that by two that's odd. Well supplied by three. Add the one, eh? It's even so divided by two same thing for a nine. Also a ten divided by two. We have another are so other formula and then even again even even still even And then finally we have anyone find Not Finally we stopped more terms but we've reached one and now we'LL see that it will It'LL start repeating and a fowling pattern. So if we use the formula for this odd number, we get three times one plus one, which is for But then once we get back to four, it will just continue to cycle for two, one for two, one and so on. And here we see the pattern once again for two one, and this will repeat So you can continue just writing those four to one four to ones and in the last four in terms of the sequence, because we are going up to forty terms, that's what's being asked and the thought the final term should be a four. So this is the sequence the first forty terms in this first case here a one is eleven. Now we'LL go to the next case Same problem, but this time a one is twenty five. So let's go ahead to the next page, and we'LL be using that same formula. So that's an odd number. So we multiply by three at once. So right that already in this time three a one plus one. Don't be seventy six. That's an even number. So recall when it's even, we divide by two, giving us thirty eight and then here. Still even so, the same formula again. Divide by two way. Haven't at number multiplied by three. Add the one around a five and that should be fifty eight. That's even divided by two odd multiply three at the one. So in this case, so we have a six. But let me back up here. Make sure my indices two and then we had twenty nine and now a seven. This should be eighty eight. That's even, you know, a ten. It's finally back to an odd. So multiply three. Add the one and thirty for We saw this number in the last problem so we could see where this will lead to. This is soon gonna lead us back to the four to one cycle. So here it's well known that a leading by two and basically just from from party here. So from previous we could see the following once we just follow the same pattern that we had in the last problem. And we could see this the same exact sequence that popped out of the previous problem. And then we went to sixteen, a four to one. And then after this point, it continued to cycle for two, one and so on. And then the last four terms, or four to one. And for and this right here, that was a forty whose and so on. So here with these numbers, we're going in order. That was a twelve. A thirteen was here, a fourteen was there and so on. And then wave reached the cycle and it kept repeating. And then it finally stopped that forty, and it happened to land on for again. So now the question is well, to make a conjecture of what will happen in the general case. So there, let's try to see what happened. So my conjecture is that regardless okay, the value of a one we'LL hear I'm assuming that it's a national number. So let's say value a one in the national numbers. The sequence will eventually reach the cycle four to one. And we saw this in the first two examples, and this is actually an open problem. No one actually has proven that this is always the case, but it looks like it's true. So here, if we go through the last two examples, we see that using the sequence. So using yes to t o find the sequence that the following terms of the sequence we see that using F allows us to reduce the on factors of the original number to reduce the other factors of its terms until it's the largest. In fact, there is just one. And then, in this case, then its terms. A m will just oscillating between four, two and one as we saw the fighter, and that's our final answer.