 ###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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# Arithmetic Series - Problem 3

Carl Horowitz ###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Finding the sum of an arithmetic series when we are defined as a summation notation. So I said arithmetic series, but before we actually go through that, we actually have to make sure that this is an arithmetic series and the easiest way to do that is just by starting to plug in values into our sigma notation, our summation notation to figure out what this pattern actually is.

So what this is telling us is starting at 0, plugging in 0, 1 minus 0 is going to give us 1. Plug in 1, 1 minus 2 times 1, 1 minus 2, -1. Plugging in 2, 1 minus 4 -3 so on and so forth.

So I actually know that I have an arithmetic sequence just by looking at the terms that come out. I actually apologize this should be an arithmetic series, this tells us we're actually adding in between these, so what we're actually trying to do is add up all these things.

So we know our formula for the sum of an arithmetic series, s sub n is equal to n over 2 a sub 1 plus a sub n. We know a sub 1 is our first term, so that's just going to be 1. What we don't know right now is a sub n, a sub n is the last term, but thinking back to how summation notations work, we go from 0, we plug in 0, 1, 2, 3 so on and so forth. The last thing we plug in is this 20, so it's pretty easy to find the last term just by plugging in 20 into here 1 minus 40 is -39, so we know that our last term then can go into this equation.

The last thing we need to figure out is how many terms we're actually adding up and this is where it gets a little bit tricky. So a lot of people just look at this and say okay highest term is 20 that's how many we're plugging in, but what you have to be careful of is we're starting at 0, so plugging in 0 gives us our first term, plugging in 1 gives us our second term, 2 gives us our third so on an so forth.

So the difference here is actually one less than the number of terms we have, so how you actually find the number of terms in a summation is the top minus the bottom plus 1. So this is actually 21 that goes in there and I'll talk about that in just a second, I just want to finish up this problem. So we have 1 plus -39 which is going to be -38 times 21 divided by 2 and our answer ends up being -399.

So I do want to take one more second to talk about how all these sigma and the number of terms work. So let's say we're dealing with the summation from j is equal to 4 to 7, I'm not even going to concern myself with what the actual problem is, I just want to worry about how many terms are.

So easy way to do this is we plug in 4, we plug in 5, we plug in 6, we plug in 7; there's four terms there. The difference here is just 3 though, 7 minus 4 is 3, so in order to account for that other term, you always have to add 1. No matter what the difference is you're always going to add 1 to figure out the number of terms.

So going from a summation notation and evaluating it, the first term is easy to find, we just plug in the bottom, the top term is easy to find we plug in the top and the number of terms is always just going to be one more than the difference.

The one thing you have to be careful of is writing out our first couple of terms just to make sure that we're actually dealing with an arithmetic series, because if we're not dealing with an arithmetic series, this formula won't hold and we'll get the wrong answer.