Okay, welcome back everyone. Last video we did some elementary examples of sigma notation. Today we're going to make it a little bit more complicated, and we're going to go over some rules, For manipulating, Slash simplifying, Or making for complicated, if you like, sigma notation. I'm going to start with our first rule, it’s just an example. Last time, remember, that we worked through that the sum from i=1 to 4 of i squared was 30. But now we're going to give you a different problem, which when you think about it is not going to be too different. The new problem is going to ask you the sum from i=1 to 4 of 3i squared. Now in fact, we could work this out directly, just as we did before. Make our little scratch work with our i and add it up. But let's work through this just enough so that we can see the pattern, see the point of this example. So if we work this out, this is equal to 3 times (1) squared + 3 times (2) squared + 3 times (3) squared + 3 times (4) squared. Now again, to the guy who's paying us to do this, we're done, you can make that into a number. Let's do that a little bit further, just for our own purposes. If we use some stuff from elementary mathematics, namely something you may remember as the distributive property, multiplication over addition, let's pull out the 3. This Is equal to 3 times [1 squared + 2 squared + 3 squared + 4 squared]. That's equal to 3 times our old friend, which we're now getting pretty bored of, as we do with old friends, the sum from i=1 to 4 of i squared. So notice the pattern here. I started off with an expression, the sum from i=1 to something. And internal here to the thing inside the sigma notation was an expression 3i squared that had a constant in it, namely 3. I can pull that constant outside and get that the sum from i=1 to 4 of 3i squared is the same thing as 3 times the same exact expression, but without the 3 sitting there. Turns out this is a general rule. Let's try one more example, if I did the sum, From r=4 to 25 of 18 r cubed, I really don't want to work that out. But I can tell you this is the same thing as 18 times the sum from r=4 to 25 of r cubed, whatever on earth that is. And this is always true. Whenever you have an expression with a constant inside the sigma, you can pull it outside, and just then evaluate it. Why is this true? There's something from elementary mathematics which you may or may not have heard called the distributive property. Namely, if you have A times (b + c) that's the same thing Ab + Ac. And all we're doing is we're taking this expression and we're doing it over and over again with the sigma notation. Okay, so that's lesson number one, that's one way of simplifying sigma notation expressions. Let's give you a new problem. The sum from i=1 to 4, and internal here we're going to have i squared plus 2i. So let's just work this out directly. So this is equal to 1 squared + 2 times (1) + 2 squared plus 2 times (2) + 3 squared + 2 times (3) + 4 squared + 2 times (4). And again we could say we're done, down tools, that's it. But let's break this up for our own purposes. Notice in this sum we have terms that look like two different things. One looks like things squared, like these ones. And one looks like two times things, like these. Let's break those up and write those separately. This is the same thing as being equal to (1 squared + 2 squared + 3 squared + 4 squared) + (2 times (1) + 2 times (2) + 2 times (3) + 2 times (4)). Let's pause for a second and remember why that's so. Basically it's so because I could add numbers in any order I want. Someone told me to add them the way I have up here, but I can add them the way I have down here, I'm just shuffling them around. Why did I do that? Well let's recognize, what's this guy here? That's equal to our old friend, the sum from i=1 to 4 of i squared. Fine, what's this guy here? That's equal to a new friend, that's equal to the sum from i=1 to 4 of 2 times i. Notice we have a rule here, I started off with the summation and then there's something inside the summation which was broken up into two different things, i squared + 2i. I can fragment that, break that up into two different summations, should I want to. Why is that true? It's true because when you add up things in any order you want. Okay, let's do our last rule here. This one seems really silly and really simple, but it makes you think about it a little bit. Let me give you the sum from k=1 to 10 of 5. Seems like a trick question. Notice the thing in the middle has no dependence on k. What do we do? This is almost convention, this is equal to, each time I evaluate something I'm going to evaluate it from k=1 to 10. But I don't do anything with k, I just write down what’s in there. This is equal to 5 + 5 + 5 +, plus, plus, dot, dot, dot, dot, + 5. How many of them? 10 of them. So this is equal to 10 times 5 = 50. The general rule here, whenever you’re summing up a constant, in this case it's 5, but it could be anything. You just add up that constant the number of times you're supposed to do. So for example, the sum from r=1 to 7 of 8 would be 8 + 8 + itself 7 times. This is the same thing as 8 times 7 or 56.