![]() ![]() Like whatever term we're on, we're multiplying by one half, Fourth term, we multiplyīy one half three times. Third term, we multiplyīy one half two times. The second term, we multiplyīy one half one time. Gonna multiply by one half? The first term, we multiplyīy one half zero times. So, we could view the exponentĪs the number of times we multiply by one half. Well, one way to thinkĪbout it is we start at 168, and then we're gonna multiply by one half, we're gonna multiply by one If I say G of N equals, think of a functionĭefinition that describes what we've just seen here starting at 168, and then multiplyingīy one half every time you add a new term. Of N, how can we define this explicitly in terms of N? And I encourage you to pause the video and think about how to do that. Times, it's often called the common ratio, times one half. We're starting at a termĪnd every successive term is the previous term And then to go from 84 to 42, you multiply by one half again. Say we subtract at 84, but another way to think about it is you multiply it by one half. If we think of it as starting at 168, and how do we go from 168 to 84? Well, one way, you could The first term is 168, second term is 84, third term is 42, and fourth term is 21,Īnd we keep going on, and on, and on. Say this is the same thing as the sequence where It is that this function, G, defines a sequence where N And we're all done.- So, this table here where you're given a bunch of Ns, N equals one, two, three, four, and we get the corresponding G of N. We know a-sub-4, the fourth term in thisĪrithmetic sequence is negative 13, so, we can now, so if this is negative 13, a-sub-5 is going to be a-sub-4, which is negative 13 minus 2, which is equal to negative 15, so the fifth term in the sequence is negative 15. Well now that we know thatĪ-sub-3 is negative 11, so this is negative 11, we could figure out a-sub-4 is negative 11 minus 2, which is equal to negative 13. Well that starts helping us out because if a-sub-2 is negative 9, if this is negative 9, then a-sub-3 is negative 9 minus 2, which is equal to negative 11. A-sub-1 is negative 7, so if this is negative 7, then a-sub-2 is negative 7 minus 2, which is equal to negative 9. We still don't know what a-sub-2 is, and so, we could write, a-sub-2 is equal to a-sub-2 minus 1, so that's a-sub-1 minus 2. A-sub-3 is going to be equal to a-sub-3 minus 1, so a-sub-2 minus 2. So we could say that a-sub-4 is equal to, well if we use the second line again, it's going to be a-sub-3, minus 2. Well, we don't know whatĪ-sub-4 is just yet, so let's try to figure that out. Let's see what we can make of this, so a-sub-5, a-sub-5 is going to be equal to, we'll use this second line right here, a-sub-5 is going to be equal to a-sub-4 minus 2. Recursive definition of our arithmetic sequence. To be a-sub-i minus 1 minus 2, so this is actually a It in terms of the previous terms, so a-sub-i is going The arithmetic sequence a-sub-i is defined by the formula a-sub-1, they gave us the first term, and they say, every other term, so a-sub-i, they're defining This arithmetic sequence is going to be 61. This is 57, and 4 plus 57 is equal to 61, so the 20th term in ![]() 3 times 19, let's see, 3 times 19 is 57, right? It's 30 plus 27, yep. This is going to be equal to 4 plus 3 times 20 minus 1 is 19. Wherever we saw an i, we put a 20, and now we can just compute We would put a 20 in, so it's going to be 4 plus 3 times 20 minus 1, so once again, just to be clear, a-sub-20, where instead of a-sub-i, A-sub-20, we just use thisĭefinition of the ith term. What is a-sub-20? And so a-sub-20 is theĢ0th term in the sequence and I encourage you to pause the video and figure out what is the 20th term? Well, we can just thinkĪbout it like this. The ith term in the sequence is going to be 4 plus 3 times i minus 1. ![]() ![]() All right, we're told that theĪrithmetic sequence a-sub-i is defined by the formula where ![]()
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