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### Course: AP®︎/College Calculus BC > Unit 6

Lesson 3: Riemann sums, summation notation, and definite integral notation- Summation notation
- Summation notation
- Worked examples: Summation notation
- Summation notation
- Riemann sums in summation notation
- Riemann sums in summation notation
- Worked example: Riemann sums in summation notation
- Riemann sums in summation notation
- Definite integral as the limit of a Riemann sum
- Definite integral as the limit of a Riemann sum
- Worked example: Rewriting definite integral as limit of Riemann sum
- Worked example: Rewriting limit of Riemann sum as definite integral
- Definite integral as the limit of a Riemann sum

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# Worked examples: Summation notation

Summation notation uses the sigma Σ symbol to represent sums with multiple terms. See some more involved examples of how we read expressions in summation notation.

## Want to join the conversation?

- Say you were given a series like:

4/3, -1, 3/4...

How would you find the formula for the terms? (like in the first example of the video?)(7 votes)- The equation for this would be that of a
**geometric sequence**and the equation would be*(4/3)*(-3/4)^(x-1)*. The way you would put it is haveat the top of the**three**`sigma`

set, and have the equation as**i equal to 1***(4/3)*(-3/4)^(x-1).*(8 votes)

- For the first example, isn't option B only a sum up to the first 3 terms?(4 votes)
- No, because we start with n=0, so we get 4 terms:

n=0, n=1, n=2, n=3(4 votes)

- If you were given something like [62, 73, 84, 95], how would you find the summation notation for this?(4 votes)
- If the difference is constant, as it is here, you can do something like making the expression the first term + index * the difference and make the boundaries 0 and the last term - 1. In your example, it would be

3

Σ (62+11i)

i=0(3 votes)

- Would we deal with Figuring out the pattern without choices ?(4 votes)
- What is the summation notation for 1/2 + 2/2 + 3/2 + 4/2 + 5/2?(2 votes)
- If we factor out 1∕2, we get

1∕2 ∙ (1 + 2 + 3 + 4 + 5) = 1∕2 ∙ ∑𝑘, as 𝑘 goes from 1 to 5.(4 votes)

- What is the summation notation for 1/2 + 4/5 + 9/10 + 16/17?(2 votes)
- Observe that the numerators form the first four positive perfect squares (in order), and the denominators are 1 more than the corresponding numerators. So the nth term of the series is (n^2)/(n^2+1), for 1<=n<=4.

So the sum of this four-term series can be written as

sum n from 1 to 4 of (n^2)/(n^2+1).(4 votes)

- so how do you actually do the math, say if the n was 1 to 500 in the situation above?(2 votes)
- So the content you are looking is arithmetic sequence. Just type that into the search bar. There is also geometric sequence.(2 votes)

- What if

(1-2+3-4+5-6+7-8+9-10)

how can we find the summation notation for this? thanks(2 votes)- Alternating positive and negative terms are common in summation notation. One way to represent this is by multiplying the terms by (-1)^i or (-1)^(i+1)(where i is the summation index). To represent your example in summation notation, we can use i*(-1)^(i+1) where the summation index is in the range [1, 10].(2 votes)

- For b, can't it not be that because it's only asking for 3 numbers for the sequence? SO wouldn't it be 2,5, and 8?(2 votes)
- The number on top the sigma is also included and substituted. So, it would be 0, 1, 2, and 3 that we would substitute. Hope this helps!(1 vote)

- That helps me a lot!(1 vote)

## Video transcript

- [Instructor] We're told consider the sum two plus five plus eight plus 11. Which expression is
equal to the sum above? And they tell us choose
all answers that apply. So like always, pause the video, and see if you can work
through this on your own. So when you look at the sum, it's clear you're starting at two. You're adding three each time. And we also are dealing with you have four total terms. Now we could try to
construct an equation here or an expression using sigma notation. But instead, what I like to
do, is look at our options. We really only have to look
at these two options here and expand them out. What sum would each of these be? Well, this is going to be the same thing as we're starting at n equals one. So this is going to be three
times, when n equals one, three times one minus one. And then plus, then
we'll go to n equals two. Three times two minus one. Then we're gonna go to n equals three. So plus three times three minus one. And then finally, we're
gonna go to n equals four. So plus three times, when n equals four, this is three times four minus one. So just to be clear, this is
what we did when n equals one. Let me write it down. N equals one. This is what we got when n equals two. This is what we got when n equals three. And this is what we got
when n is equal to four. And we stopped at n equals four 'cause it tells us right over there, we start at n equals one and we go all of the way to n equals four. So what does this equal? So let's see, three times
one minus one is two. So this is looking good so far. Three times two minus one, that's six minus one. That is five. So still looking good. Three times three minus
one, that's nine minus one. Once again, still looking good. Three times four minus one, that is 11. So we like this choice, so I would definitely select this one. Now let's do the same thing over here. When n is equal to zero, it's going to be two
plus three times zero. So that's just going to be two. And then plus, when n is equal to one, it's going to be two plus three times one, which is five. This is starting to look good. Now when n is equal to,
this was n equals zero, this was n equals one, so now we're at n equals two. So at n equals two, two plus three times two is two plus six, which is eight. And this makes sense. Every time we increase n by one, we are adding another three, which is consistent
with what we saw there. We start at two, when n equals zero, this is just, that three n is just zero. So you start at two, and then you keep, every time you increase n by one, you are just adding three again. So finally, when n is equal to three, two plus three times three is two plus, or is 11, I should say. And so this also is exactly the same sum. So I feel good about both of these. Let's do one more example. So we're here, we're given the sum, and we're saying choose one answer. Which of these is equivalent
to this sum right over here? Well, like we did before,
let's just expand it out. And what's different here is
that we just have a variable, but that shouldn't make it
too much more difficult. So let's do the situation,
and I'll write it out. Let's do the situation
when n is equal to one. When n is equal to one, it's
gonna be k over one plus one. K over one plus one,
and I'll write it out. This is n is equal to one. And then plus, when n is equal to two, it's gonna be k over two plus one. This is n is equal to two. And then we're gonna keep going. When n is equal to three, it's gonna be k over three plus one. That's n is equal to three. And then plus, finally, 'cause we stop right over
here at n equals four, when n equals four, it's going
to be k over four plus one. That's when n is equal to four. So this is all going to be equal, I'll just write it over here, this is equal to k over two, plus k over three, plus k over four, plus k over five, which is exactly this choice right over, right over here. And actually, if I had looked
at the choices ahead of time, I might have even been
able to save even more time by just saying, well, look, actually, if you just try
to compute the first term, when n is equal to one,
it would be k over two. Well, only this one is
starting with a k over two. This, this one has no k's
here, which is sketchy. They're trying to look at the error where you try to replace the
k with the number as well, not just the n. So that's what they're trying to do here. Here they're trying to, let's see, well, this isn't as obvious
what they're even trying to do, right over here where they
put the k in the denominator. And here if you swapped the n and the k's, then you would've gotten
this thing right over here. So we definitely feel good about choice A.