Sure this second method, right over here where we'd come You might be a little bit,Ī toss up on which method you want to use, but for So this is equal to negative 1/8, times two to the third power. Is equal to negative 1/8, times two to the four, minus one. Using this explicit formula, we could say a sub four, So we want to find theįourth term in the sequence, we could just say well, We're going to take our initial term, and multiply it by two, once. Based on this formula, a sub two would be negative 1/8, times two to the two minus one. A sub one, based on this formula, a sub one would be negative 1/8, times two to the one minus one. We're going to multiply itīy two, i minus one times. We could explicitly write it as a sub i is going to be equal to our So we could explicitly, this is a recursive definitionįor our geometric series. We know each successive term is two times the term before it. Another way to think about it is, look, we have our initial term. Two times negative 1/2, which is going to beĮqual to negative one. Is equal to negative 2/4, or negative 1/2. It's going to be two times negative 1/8, which is equal to negative 1/4. Lucky for us, we know thatĪ sub one is negative 1/8. Then we go back to this formula again, and say a sub two is going Go and use this formula, is going to be equal A sub four is going to beĮqual to two times a sub three. We could say that a sub four, well that's going to be What is a sub four, theįourth term in the sequence? Pause the video, and see That is defined as being, so a sub i is going to be two Where the first term, a sub one is equal to negative 1/8, and then every term after Geometric sequence a sub i, is defined by the formula So I can reuse most of my equation from my simple example: a(i) = a(1) ∙ (2) ^ (i - 1) So for Sal's example, the terms are messier and we start out knowing only the first value and the multiplier, and the important information that it follows the rules for a geometric sequence.Įach term is 2 times the previous. If we want to find the 4th term, here is how we calculate it: In this simplified case I showed above, a(1) is 3 This is sometimes called the explicit formula, because you can generate any term if you know the first value and multiplier (common ratio). If you don't adjust the exponent by one, you will find terms that are in the wrong location. You can write a quick, general formula from this for all geometric sequences:įirst value x multiplier raised to number of the term, minus one If you have an original number of 3, your term numbers i would look like this top row. Another way to think of it is that every time you need a new term, you multiply by 2.
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