## 1151 FF: Sigma Notation and Sum Formulas

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.