Let’s work through a few examples of sigma

notation and sum formulas. The first example, number 1, is the sum from K equals 1 to 53

of K. As I read that, I kind of said what that notation means. This symbol that sort

of looks like an E maybe is called a sigma. Then you notice we have a subscript and a

superscript. Then we have K which in this case is what we’re actually summing. If I

were to do this by hand, what this would be is I would be plugging in these values from

1 to 53 for K and summing them up. This would be like 1 plus 2 plus 3 plus so on, all the

way up to 53. Now of course I could do that by hand and get the answer, but the point

of these sum formula’s we just looked at, is to have a more efficient way of calculating

large sums like these. Let me write down the relevant formula here. We just saw that the

sum from 1 to N of K is N times N plus 1, over 2. Let’s apply that formula to the problem

that we have. If you kind of line up our problem with the formula, our N is 53. We’re summing

from K equals 1 to 53. I’m just going to plug into my formula. This is going to equal 53

times 54 divided by 2. If you put that into your calculator and multiply it out, you get

1431. Let’s do one that’s a little bit more complicated. We could have something that

looks like this. The sum from K equals 1 to 21 of 2 K squared minus 2 K minus 5. If I

were going to do this out by hand, what would I do? I would start by plugging in K equals

1. I would get whatever that value is. Plug in K equals 2, get whatever that value is.

Plug in all the numbers all the way up to 21 into this expression, 2 K squared minus

2 K minus 5. Figure out what that is and then add up all those values that I get. That’s

a lot of work. Once again we can use the sum formulas we just saw to calculate an answer

to this a little more efficiently. Once again let me write the relevant formulas out to

the side. We see we have a K squared. We are going to need the formula for K squared. It’s

the most messy of the formulas. It looks like this: N times N plus 1 times 2 N plus 1 over

6. I’ll write down the formula for K one more time. The sum of K. That was N times N plus

1 over 2. Then we also have this constant term. There is also a formula for the sum

of a constant which we labeled C, which was C times N. Let’s see if we can apply these

formulas. What you can do to help yourself out is you can actually split this up in the

same way that you would split up a derivative or an integral. For example, I can pull constants

out front, constant multiples. I can split up these differences. If I do that, it’s going

to look like this. I pull out my constant multiple of 2 out front, and then I have the

sum of K squared, minus another constant multiple of 2. Then I have the sum of K minus the sum

of 5. I’ve split this up. Now I can use the appropriate formula on each piece. On this

first piece I have this 2 out front which I’ll keep there for now. I’m going to use

the K squared formula. Now my N in this case is 21. I want to plug in 21 for N. I get 21

times 22, 2 N plus 1, 2 times 21 is 42 plus 1 is 43. That’s all over 6. Minus 2 again.

Now I need my K formula. Once again with N equals 21, this gives me 21 times 22 over

2. Then minus my formula for a constant is just that constant times N. That’s going to

5 times 21. Now we need to calculate these values. If you plug this first once into your

calculator, you should get 6622 assuming I did that correctly, minus, well this should

just give us 21 times 22 because the 2’s cancel out. If you actually calculate that, you get

462. Then minus 5 times 21 which is 105. If you put this into your calculator as your

final answer you should get 6055. Hopefully this gives you a good sense of how to work

with something like this. Now you need to be a little careful. Let me go back to this

original one that we did. We summed K from K equals 1 to 53. Now notice that my sum formula

specifically starts from K equals 1 and goes to N. You can only use this formula in this

form if your K starts at 1. What do I mean by that? What if, instead, the sum from K

equals to 53 of K? I can’t use just this formula to solve this now because I’m not starting

at K equals 1. I can still use the formula though, I just have to use a little bit of

extra thought. What’s happening in this problem as opposed to the one I did? I’m leaving off

those first two terms: K is 1 and K is 2. What I can actually do in this case is I can

use my original answer, which would be the sum of all the original 53 terms, but then

subtract off the first 2 terms. Which would be plugging in K equals 1 and K equals 2.

In this case it would be minus 1, minus 2. Minus 3. What would that be? 1428 I guess.

Just be careful in the way you use these formulas and just be thoughtful. They are very useful

formulas so I would highly recommend memorizing these.