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## Series basics challenge

Current time:0:00Total duration:6:16

# Sum of n squares (part 1)

## Video transcript

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.