What I want to do in this video is come up with an expression
for finding the sum from i equals 0 to n of i squared.
So if I were to expand this out, this
is equal to 0 squared plus 1 squared
plus 2 squared plus 3 squared.
And we're going to keep on going all the way to n squared.
So my goal is to find some type of a function
that you give me the n and I will
find the sum from 0 squared, 1 squared,
2 squared, all the way to n squared.
And so you can imagine that'd be useful,
because this might be OK if n is reasonably small.
But if n is a big number, this is
going to take you forever to do.
So let's first study this.
Let's study what the input and the output of this function
needs to be.
So the input is going to be our n.
So here we're starting with-- so n can go from 0 all the way--
and we'll just try up a bunch of values.
So n could be 0.
We could go from 0 all the way to 0.
n could be 1.
n could be 2.
n could be 3.
And we could just keep on going on and on and on.
But I'll just stop there for now.
Actually, let's just go to 4, just for fun.
And now, for each of these, let's see
what the output of our function should be.
The output of the function should be this thing.
It should be the sum from i equals 0 to n of i squared.
So when n is 0-- well, that's just going to be 0 squared.
We'd just stop right over there.
So that's just 0.
When n is 1, it's 0 squared plus 1 squared.
So that is 1.
When n is 2, it's 0 squared plus 1 squared plus 2 squared.
So that's 1 plus 4, which is 5.
When n is 3, now we go all the way to 3.
So it's going to be 1 plus 4, which is 5, plus 9.
So 5 plus 9 is 14.
And then when n is 4, we're going to add the 16, 4 squared,
to this.
So this gets us to 30.
And of course, we could keep going on and on and on.
So let's study this a little bit to think
about what type of a function that, for each of these inputs,
might give us this type of an output.
So let's first look at the difference between these terms.
So the difference here is 1.
The difference here is 4.
And this is obvious.
We added 1 here.
We added 2 squared here.
We added 3 squared, or 9, here.
We added 4 squared, or 16, here.
And the reason why I'm doing this
is if this was a linear function,
then the difference between successive terms
would be the same.
Now, if this is a quadratic function,
then the differences between the differences would be the same.
Let's see if that's the case.
So the difference here is 1.
The difference here is 4.
So the difference between those is 3.
The difference here is 5.
The difference here is 7.
So even the difference of the differences is increasing.
But if this is a cubic function, then the differences
of the difference of the difference should be constant.
So let's see if that's the case.
And you'll appreciate this even more
when you start learning calculus.
So let's see.
The difference between 3 and 5 is 2.
The difference between 5 and 7 is 2.
And so we keep having a constant shift of 2.
So the fact that the difference of the difference
of the difference is fixed tells us
that we should be able to express this
as some type of a cubic function.
So this we could write as this should
be equal to some function in terms of n.
And we could write it as An to the third plus Bn squared
plus C times n plus D. And now we
can just use what the inputs are and the outputs are of these
to solve for A, B, C, and D. And I encourage you to do that.
Well, let's first think about when n is equal to 0.
When n is equal to 0, this function evaluates to D.
So this function evaluates to D. But that function
needs to evaluate to 0, so D needs to be 0.
So I'm just trying to fix these letters here
to get the right outputs.
So when n is 0, this expression evaluates to D.
And it needs to evaluate to 0, so D needs to be equal to 0.
So D is equal to 0, or we could just ignore it.
So that helps us a little bit.
We know from this data point we're
able to whittle it down to it having
this form right over here.
And so now we can take each of these inputs
and figure out what their corresponding output is.
So let's do that-- I'll do that over here.
So when n is 1, this thing evaluates
to-- let me do this in a new color-- this thing
evaluates to A times 1 to the third power, which is just 1,
plus B times 1 squared, which is just 1, plus C times 1, which
is just C. And this needs to be equal to 1.
Now, when n is 2, we have A times n to the third.
So that's 8A plus 2 squared is 4,
plus 4B plus 2C needs to be equal to 5.
And I need to set up three equations
if I want to solve for three unknowns.
And so let's go to 3.
So A times 3 to the third power, so that's going to be 27A,
plus 9B plus 3C is going to be equal to 14.
So I've set up three equations in three unknowns.
Now I just have to solve these for A, B, and C.
And I will have a generalized formula for finding this sum
right over here, the sum of the first n numbers squared,
I guess you could call it.
So what I want to do now is I'm going to stop this video.
I encourage you to try to solve the simultaneous equation
on your own.
In the next video, I'll actually go and solve it.