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# Worked example: arithmetic series (recursive formula)

## Video transcript

- [Voiceover] So I'm going
to recursively define an arithmetic sequence. So we're going to say that
they ith term of the sequence is equal to the i minus oneth term of the sequence plus 11. So each term is going to be 11 more than the term before it. Now we have to establish a base case here, and so we're going to
say that the first term of our arithmetic sequence is going to be equal to four. So given this recursive definition of our arithmetic
sequence right over here, what I challenge you to do is to find the sum of the first 650 terms of the sequence. Let me write that down. Find the sum of first 650 terms of the sequence, of this arithmetic sequence that we have just defined. And like always, pause the video and see if you can work that out. All right, so how can we think about this? Well, in many videos we give our intuition for the sum of an arithmetic sequence, and we came up with a formula for evaluating a sum of
an arithmetic sequence, which we call an arithmetic series, and that sum of the first n terms is going to be the first term plus the last term over two, so really the average of
the first and last terms, times the number of terms we have. And this is only the case
when it's an arithmetic series where each term that we're adding is a fixed amount larger or less than the term before it, or that
we have a fixed difference. So what about this one right over here? What is the first and the last term going to be, and what is our n? Well we know that n is 650, we know that n is 650, and we know what the first term is going to be. The first terms is going to be four. We need to figure out
what the nth term is, or we need to figure out
what the 650th term is going to be. Let's think about this a little bit. So this is going to be,
if we're taking the sum, it's going to be four plus the next term, the second term, so if a sub two is going
to be a sub one plus 11. So it's going to be four
plus 11, which is 15. We're going to add 11,
which is going to get us to 26, and we're going to keep adding 11. Now how many times are we going to add 11? Well to get to the second term, we add 11 once. To get to the third term, we add 11 twice. So to get to the 650th term, so this is a sub 650, a sub 650, we're going to have to add 11, look, to get to the second term, we added 11 once, third term, add 11 twice. So to get to the 650th term, we are going to add 11, we are going to add 11
650 minus one times, or 649 times. Notice, to get to each term, to get to the first term, you added one minus one, you had an 11 one minus one times, you added 11 zero times. You started with the four, didn't add 11 at all. Then the second term, you added 11 once. Third term, you added 11 twice. Fourth term, you added 11 three times. 650th term, you added 11 649 times. And so if you add 11 649 times, what do you get? So four plus 649 times 11 is going to be equal to, I'll get my calculator out for this, so this is going to be equal to 649 times 11 is equal to, now plus four is equal to 7,143, 7,143. So that's the 650th term. 7, 143. And so now we can just evaluate this. So I'll get the calculator out for that. So we have 7,143 plus four plus the first term plus four is equal to that. We're going to divide by two. So divide by two gets us 3,573.5. We're going to multiply that times 650. That's how many terms we have. Times 650 is equal to, that's a pretty large number, is going to be equal to 2,322,775. 2,322,775. I already forgot it. I have trouble remembering things when I take it off my screen. All right, 2,322,775. I'm glad I had a calculator in hand for that one, but you could do it by hand. I always encourage you to do it. It never hurts to practice the arithmetic.