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.