![]() Another way you could think about it is, well, let's use our exponent And you can think of it in other ways, you could write thisĪs G of N is equal to, let's see, one way you could write it, as, you could write it as 168,Īnd I'm just algebraically manipulating it over Nice explicit definition for this geometric series. You're gonna multiply by one half twice, and you see that right over there. If N is two, well, two minus one, you're gonna multiplyīy one half one time, which you see right over here, N is three, you're gonna multiply by one half twice. If N is equal to one, you're going to have one minus one, that's just gonna be zero. 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 Converting is usually less work.- 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. Thankfully, you can convert an iterative formula to an explicit formula for arithmetic sequences. In the explicit formula "d(n-1)" means "the common difference times (n-1), where n is the integer ID of term's location in the sequence." ![]() In the iterative formula, "a(n-1)" means "the value of the (n-1)th term in the sequence", this is not "a times (n-1)." ![]() Even though they both find the same thing, they each work differently-they're NOT the same form. A + B(n-1) is the standard form because it gives us two useful pieces of information without needing to manipulate the formula (the starting term A, and the common difference B).Īn explicit formula isn't another name for an iterative formula. M + Bn and A + B(n-1) are both equivalent explicit formulas for arithmetic sequences. So the equation becomes y=1x^2+0x+1, or y=x^2+1ītw you can check (4,17) to make sure it's right Substitute a and b into 2=a+b+c: 2=1+0+c, c=1 Then subtract the 2 equations just produced: Solve this using any method, but i'll use elimination: The function is y=ax^2+bx+c, so plug in each point to solve for a, b, and c. Let x=the position of the term in the sequence Since the sequence is quadratic, you only need 3 terms. ![]() that means the sequence is quadratic/power of 2. However, you might notice that the differences of the differences between the numbers are equal (5-3=2, 7-5=2). This isn't an arithmetic ("linear") sequence because the differences between the numbers are different (5-2=3, 10-5=5, 17-10=7) ![]()
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