Arithmetic series formula
Let's write an arithmetic sequence in general terms. So we can start with some number a. And then we can keep adding d to it. And that number that we keep adding, which could be a positive or a negative number, we call our common difference. So the second term in our sequence will be a plus d. The third term in our sequence will be a plus 2d. So we keep adding d all the way to the n-th term in our sequence. And you already see here that in our first term, we added d zero times. Our second term, we added d once. In our third term, we added d twice. So you see, whatever the index of the term is, we're adding d one less than that many times. So if we go all the way to the n-th term, we're going to add d one less than n times. So it's going to be n minus 1 times d. Fair enough. And let me write that. This right over here is our n-th term. Now what I want to do is think about what the sum of this arithmetic sequence would be. And the sum of an arithmetic sequence we call an arithmetic series. So let me write that in yellow. Color changing is sometimes difficult. So the arithmetic series is just the sum of an arithmetic sequence. So let's call my arithmetic series s sub n. And let's say it's going to be the sum of these terms, so it's going to be a plus d, plus a plus 2d, plus all the way to adding the n-th term, which is a plus n minus 1 times d. Now I'm going to do the same trick that I did when I did the most basic arithmetic sequence. I'm going to add this to itself, but I'm going to swap the order in which I write this sum. So s sub n I can write as this, but I'm going to write it in reverse order. I'm going to write the last term first. So the n-th term is a plus n minus 1 times d. Then the second to last term is going to be a plus n minus 2 times d. The third to last is going to be a plus n minus 3 times d. And we're going to go all the way down to the first term, which is just a. Now let's add these two equations. We are going to get, on the left hand side, s sub n plus s sub n. You're going to get 2 times s sub n. Well, what's the sum of these two first terms right over here? I'm going to have a plus a plus n minus 1 times d. So it's going to be 2a plus n minus 1 times d. Now let's add both of these second terms. So if I were to add both of these second terms, what do I get it? I'm going to get 2a plus 2a. And what's d plus n minus 2 times d? So you could view it several ways. Let me write this over here. What is d plus n minus 2 times d? Well, this is just the same thing as 1d plus n minus 2 times d. And so you could just add the coefficients. So this is going to be n minus 2 plus 1 times d, which is equal to n minus 1 times d. So the second term also becomes 2a plus n minus 1 times d. Now let's add the third term. I'll do it in green. The third terms, I should say. And I think you're going to see a pattern here. It's 2a plus 2a. And if I have 2 plus n minus 3 of something and then I add 2, I'm going to have n minus one of that something. So plus n minus 1 times d. And you're going to keep doing that all the way until your n-th pair of terms, all the way until you add these two characters over here, which is just 2a plus n minus 1 times d. So you have this 2a plus n minus 1 d being added over and over again. And how many times are you doing that? Well, you had n pairs of terms when you were adding these two equations. In each of them, you had n terms. This is the first term, this is the second term, this is the third term, all the way to the n-th term. So I can rewrite 2 times the sum 2 times s sub n is going to be n times this quantity. It's going to be n times 2a plus n minus 1 times d. And then if we want to solve for s sub n, you just divide both sides by 2. And you get s sub n is equal to, and we get ourselves a little bit of a drum roll here, n times 2a plus n minus 1 times d. All of that over 2. Now, we've come up with a general formula, just a function of what our first term is, what our common difference is, and how many terms we're adding up. And so this is the generalized sum of an arithmetic sequence, which we call an arithmetic series. But now, let's ask ourselves this question. This is hard to remember. The n times 2a plus n minus 1 times d over 2. But in the last video, when I did a more concrete example, I said well, it looks like the sum of an arithmetic sequence could be written as perhaps the average of the first term a1 plus an. The average of the first term and the last term times the number of terms that you have. So is this actually the case? Do these two things gel? Because this is very easy to remember-- the average of the first and the last terms multiplied by the number of terms you had and actually makes intuitive sense, because you're just increasing by the same amount every time. So let's just average the first and the last term and then multiply times the number of terms we have. Well, all we have to do is rewrite this a little bit to see that it is indeed the exact same thing as this over here. So all we have to do is break out the a. So let me rewrite it. So, this could be rewritten as s sub n is equal to n times a plus a plus n minus 1 times d. I just broke up this 2a into an a plus a. All of that over 2. And you see, based on how we defined this thing, our first term a1 is a. And then our last term, a sub n, is a plus n minus 1 times d. So this whole business right over here really is the average of the first and last terms. I got my first term, adding it to my last term, dividing it by 2. And then I'm multiplying by the number of terms we have. And that's true for any arithmetic sequence, as we've just shown here.