Everyone, welcome back. So in this example problem, we're given the first four terms of a sequence. We've got these four terms separated by commas, and we're going to use this sequence of numbers to write the general formula. So, in other words, we're going to look at the patterns of numbers and figure out if we can find an equation for \( a_n \), the general or nth term. And that's just going to be some equation, and then we're going to use this equation that we find to find the 15th term of the sequence. So rather than having to continue out the pattern to the 15th term, we're going to be able to calculate this very quickly just by plugging 15 in for \( n \) once we figure out our formula. Alright? So let's go ahead and get started here.
So if you look at this, remember that when we look at the sequence of numbers, we're going to look at the patterns of numbers to figure out what's going to be in our formula. There are a couple of things. For example, if you notice that numbers are increasing by the same amounts for each term or if you see alternating signs, things like that. Let's take a look at our sequence. We have \( \frac{1}{1 \times 2} \), \( \frac{1}{2 \times 3} \), \( \frac{1}{3 \times 4} \), \( \frac{1}{4 \times 5} \). So this is a fraction. And remember, whenever we have fractions, that means that your general formula is going to have a fraction. So that's a good place to start.
Now we're going to take a look at the top and the bottom because those patterns will usually be different. For example, notice how the numerator, the tops of each one of the fractions, no matter what the term is, is always 1. What that tells us is that this top doesn't depend on what the index is. It has nothing to do with \( n \). So whenever you see ones on the top of each one of your numbers, that just means that there's going to be a one in your general formula because this thing, no matter what \( n \) is, is always just going to be 1. Alright? So that one's pretty easy, the numerator.
Let's take a look at the denominator because it's a little bit different. Right? You can notice here that we're going to actually have two numbers that are always being multiplied in the denominator: \( 1 \times 2 \), \( 2 \times 3 \), \( 3 \times 4 \), \( 4 \times 5 \). Also, notice how the numbers are always different each time. In other words, we have \( 1 \times 2 \), \( 2 \times 3 \). Right? So the numbers are constantly changing across the denominator. So what that tells us is that we're definitely going to have some kind of an \( n \) in our denominator. So let's take a look at the first term in each of the denominators because this one starts at 1 and then the next, in the next term, the first number is a 2. In the next term, the first number is a 3. In the next term, the first number is a 4. So in other words, between each one of the first terms of the denominator, we're increasing by the same amount each time. This is actually like plus 1. Now remember, whenever we have numbers that increase by the same amount each time, then that's going to be some kind of a multiple of \( n \). And when it increases by 1, that's just going to be \( n \). And notice how over here, what happens is that the first term, the index of 1, corresponds with the first number being 1. An index of 2, the second term means that number is a 2. The 3rd term, that first number is a 3. And the 4th number, that first number is a 4. So that means that this is actually literally just \( n \). Right? Because as the index changes as we go from 1, 2, 3, 4, then this number is going to be 1, 2, 3, 4. Alright? That first number is definitely going to be \( n \) in our general formula.
Now let's take a look at the second number over here. The second number in the in the denominator is 2 and then 3 and then 4 and then 5. So what you'll notice here is that between the terms, the second number in the denominator also increases by 1. So does that just mean that we're just going to have to multiply \( n \) times \( n \)? Well, if you think about it, not exactly because \( \frac{1}{n \times n} \) means that if you were to plug in an index of 1, this would be \( \frac{1}{1 \times 1} \). And that's not what the first term tells us. It's \( \frac{1}{1 \times 2} \). So notice how it happens is that this index over here of 1, this number is always one higher than that index. Right? So here for the second term, we have the number 3. For the 3rd term, the index of 3, that second number is a 4. So it's definitely going to be \( n \), but remember that sometimes you're going to have to adjust your formula by adding, subtracting, or multiplying, and dividing constants. And in this case, because everything is shifted up by 1, then in our formula, we have to include \( n + 1 \). So now what happens is if you plug in \( n = 1 \), you're going to get \( \frac{1}{1 \times (1+1)} \), which is going to be \( \frac{1}{2} \). So that's how we adjust for that number, kind of starting at some number that isn't 1. So we're going to have to raise this by 1. So if you plug in, so, for example, just \( n = 1 \), which you're going to see, this is going to be \( \frac{1}{1 \times (1+1)} \). So this is going to be 2. If you plug in an index of 2, remember, this is just going to the first term is going to be 2, and then the second term is going to be \( 2+1 \). So this is 2+1, and this is going to be \( \frac{1}{2 \times 3} \). So notice how if you keep sort of, if you keep actually extending out the numbers, you'll see that the \( a_1 \), \( a_2 \), \( a_3 \), \( a_4 \) are exactly match what's going on in the sequence. Alright? So this definitely is the general formula for our sequence over here. It's \( \frac{1}{n \times (n+1)} \). Sometimes you might see this in parentheses or something like this if there's if there's, you know, multiplication that's going on. So that's really what our general formula is. So how do we use this now to find the 15th number in the sequence? Well, it's pretty straightforward. Basically, what the what this \( a_n \) formula tells us is that to find out the 15th term, we're just going to do 15 as our \( n \). So in other words, this is going to be 15 times, and this is going to be 15 plus 1. Alright? So this is just going to be the next number up. So this is just going to be \( \frac{1}{15 \times 16} \). And if you actually multiply this out in your calculators, which you should see is that this is about 0.0042 repeating. It's going to be 416, but it's just going to round to 0.0042. Alright? So that is the 15th term in your sequence, and that's the general formula. So thanks for watching, and I'll see you in the next one.