Main content

## Algebra (all content)

### Course: Algebra (all content) > Unit 18

Lesson 6: Advanced sigma notation- Arithmetic series in sigma notation
- Arithmetic series in sigma notation
- Finite geometric series in sigma notation
- Finite geometric series in sigma notation
- Evaluating series using the formula for the sum of n squares
- Partial sums intro
- Partial sums: formula for nth term from partial sum
- Partial sums: term value from partial sum
- Partial sums intro

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Evaluating series using the formula for the sum of n squares

Using properties of sigma notation to rewrite an elaborate sum as a combination of simpler sums, which we know the formula for. Created by Sal Khan.

## Want to join the conversation?

- Am I just being completely foolish or can you not find the sum of this by finding the area under the curve from 1 to 7?(24 votes)
- This reply is a little late, but hopefully someone will find it useful... You are not being foolish at all! The area under the curve described by the formula given for the series will give the same answer. Calculus and series are related in that way. Sal should give more detail on this in lectures covering Taylor and Maclaurin series. So your question is actually a very good one.(31 votes)

- At6:30Sal talks about looking up the equation on khan academy of the sum of n^2 but where would we look it up at? Is it in the form of a video or a FAQ?(12 votes)
- You can find the explanation for that formula used at6:30in the videos "Finding the sum of n squares part1 and part2"(11 votes)

- @3:45, Sal gives an example as a sum of fractions. Why? Shouldn't it be 1^2 +2^2+....+7^2?(4 votes)
- He is using a formula to find the sum, rather than finding each value and adding them one by one.(7 votes)

- why doesn't n((a1+a2)/2) work for all of sigma notations?(6 votes)
- The formula n(a1+an)/2 can only be used to find the sum of an arithmetic series with n terms. Notice here that a1 is the first term of the series, and an is the last term. Hence, it cannot be applied to find the sum of squared terms.(4 votes)

- Waait... where does Sal get the 7/6 at @4:10?(6 votes)
- He gets it from the equation at3:50. You can derive it yourself or find derivations for it.(3 votes)

- I don't get what Sal did at4:13. Anybody know?(2 votes)
- He's using a simplified formula to the sum of n², he has 3 videos detailing where that formula comes from, in order they are:

https://www.khanacademy.org/math/integral-calculus/sequences_series_approx_calc/calculus-series/v/sum-of-n-squares-1

https://www.khanacademy.org/math/integral-calculus/sequences_series_approx_calc/calculus-series/v/sum-n-squares-2

https://www.khanacademy.org/math/integral-calculus/sequences_series_approx_calc/calculus-series/v/alternate-sum-of-n-squares-formula(8 votes)

- At4:10, how did you get both of those formulas for evaluating the sum?(3 votes)
- https://www.khanacademy.org/math/integral-calculus/sequences-series-approx-calc/calculus-series/v/sum-of-n-squares-1

this is the link to the derivation of above formulae(3 votes)

- In1:55it says that 4 goes to 1 to 7 or the sum of 4 from 1 to 7 (n=1)...since n=1 and there is no variable n in it...so why is the answer 7*4=28 or 4+4+4+4+4+4+4=28 ....why isn't the answer just 4 since there is no "n" variable..?(2 votes)
- We have to add 4 to our sum 7 times because we're indexing from n=1 to n=7. The fact that n is not in the summand has no bearing on how we interpret the sigma symbol; it just means that the summand is constant for all indices rather than being a variable term that depends on the index. Hope that makes sense.

Since you seem to be familiar with JavaScript, you might like to think of it this way:`// This is SUM(n=1 => 7) [ 4 ]`

var sum = 0;

for (var i = 1; i <= 7; i++) {

sum += 4;

}

// As you can see, we add 4 to the sum 7 times.

// ***************

// This is SUM(n=1 => 7) [ 4*n ]

var sum2 = 0;

for (var n = 1; n <= 7; n++) {

sum2 += 4*n;

}

// The summand depends on n in this sum.(4 votes)

- how do I find those formulas?(2 votes)
- sorry for the noob question--i understand how sal got the equation for the sum of n^2 but can't one also use the equation for geometric sequences

Sn=(a-ar^n+1)/1-r ? When I plug in 1 for a and 2 for r (common ratio), I got 256 using this equation-- clearly I am doing something wrong. Would appreciate any help

THanks(2 votes)

## Video transcript

Let's try to find the sum
of this right over here, or let's try to evaluate this
expression right over here. So we're evaluating what
this sum turns out to be. Now, there's a bunch
of ways to do this. You could literally just
do it by brute force. You could say, well,
what does this equal when n equals 1, when n equals
2, all the way to n equals 7? And that would be
completely legitimate. But I view this
is an opportunity to look at some properties
of sigma notation. So let's split this out. So the first thing
that you might say is, well, look, if
I were to sum up all of these when n equals 1
and n equals 2, all the way to n equals 7, it's reasonable
that this is going to be the same thing as the
sum of 3 n squared from n equals 1 to 7 plus the
sum from n equals 1 to 7 of 2n squared-- of
2n, I should say, that right over there, plus the
sum from n equals 1 to 7 of 4. And if you find this a
little bit confusing, I encourage you to expand
to both of these things out and realize that when
you rearrange the terms, that you will get
these two things. I'm not doing a
rigorous proof here, but hopefully, if you
were to expand this out, you'd see that this is
not an unreasonable thing to claim right over here. Now, out of all of
these, this last piece is pretty easy to evaluate. When n equals 1, this
thing is equal to 4. When n equals 2, this
thing is equal to 4. When n equals 3, this
thing is equal to 4. So you're essentially going
to take seven 4's and add them together. So this is essentially just
going to evaluate to 7 times 4, or 28. Now let's look at this
piece right over here. Now, once again, we can
just do it by brute force. 2 times 1 is 2 plus
2 times 2 is 4. So you're going to
essentially do the first seven multiples of 2 is one way
you could think about this. Or, if we were to expand
it out-- actually, let me expand it out. This is going to be 2 plus
4 plus 6, all the way to, when this is 7,
all the way to 14. You could factor out a 2. And so this is going to
become 2 times 1 plus 2 plus 3 all the way to 7. And so you can rewrite
this piece right over here as 2 times the sum--
so we're essentially just factoring out the 2--
2 times the sum, which is the sum from n
equals 1 to 7 of n. So this is this piece. We still have this 28
that we have to add. So we have this 28. And we draw the
parentheses so you don't think that the 28 is
part of this right over here. And now we can do the
same thing with this. 3 times n-- we're taking from
n equals 1 to 7 of 3 n squared. Doing the same exact thing
as we just did in magenta, this is going to be equal
to 3 times the sum from n equals 1 to 7 of n squared. We're essentially
factoring out the 3. We're factoring out the 2. n squared. And once again, we can
put parentheses just to clarify things. Now, at this point, there
are formulas to evaluate each of these things. There's a formula to evaluate
this thing right over here. There's a formula to evaluate
this thing over here. And you can look them up. And actually, I'll
give you the formulas, in case you're curious. This formula, one
expression of this formula is that this is going to
be n to the third over 3 plus n squared over
2 plus n over 6. That's one formula for that. And one formula for this piece
right over here, going from n equals 1 to 7-- sorry. Let me make it clear. This n is actually what your
terminal value should be. So this should be 7 to
the third power over 3-- so it's not this n. I was just mindlessly using the
formula-- 7 to the third over 3 plus 7 squared over 2 plus 7/6. So that's this sum. And this sum, you could
view it as the average of the first and the last terms. So the first term is 1. The last term is 7. So take their average
and then multiply it times the number
of terms you have. So times-- you have 7 terms. So what is this middle
one going to evaluate to? Well, 1 times-- and of course,
we have this 2 out front. This green is just this
part right over here. So you have 2 times this. And over here, you have 3 times
this business right over here. So if we evaluate this
one, 2 times-- let's see. 1 plus 7 is 8,
divided by 2 is 4. 4 times 2 is 8. Times 7, it's 56. So that becomes 56. Now, this-- let's see. This is actually-- well, we
could evaluate this if we want. And I guess we could
take out a calculator if we wanted to figure
out 7 to the third power. Actually, let's just do
that just to save time here. So let's calculate. So we have 7 to the
third power divided by 3 plus 7 squared divided
by 2 plus 7 divided by 6 gives us a drum roll of 140. So this is going to
be equal to 3 times 140-- let me do it in that
color-- 3 times 140 plus 56 plus 28. And since we get our calculator
out, let's just use it. So let's see. 140 times 3 is 420,
of course, plus 56, plus 28-- we deserve a drum
roll now-- gets us to 504. So this sum right over
here is equal to 504. And once again, multiple
ways you can do it. But it's nice to know
that there are these ways to break down the problem. And there are all
these formulas. Now, I encourage you
to look at the formulas to see how this is actually
derived and proved. I'm not a big fan
of just saying, oh, there's a formula for this. You just apply it. The formula here is
whatever this terminal value is to the third power over
3 plus that squared over 2, plus that over 6. I encourage you to
look up on our site, on Khan Academy, the formula
for the sum of n squares, and it'll tell you where
this is derived from. And also, the formula for the
sum of an arithmetic series, and it'll tell you where
this is derived from.