[music] What do I get if I take one cubed
plus two cubed plus three cubed plus so on
plus K cubed?
This is a truly remarkable fact it's
called Nicomachus' Theorem, and it says
that the sum of the first K perfect cubes
is the square of just the sum of the first
K whole numbers, and it's literally a
remarkable fact that these two things are,
are equal, right?
I mean, you know, it just doesn't look
like this should be true at all, but it
is.
Now, once you believe something like this
you could rewrite this sum of cubes
formula in maybe a slightly easier way.
Right, once you've got the sum of cubes is
the square of just the sum of the first k
whole numbers, while we already know a
formula for some of the first k whole
numbers.
It's k times k plus 1 over 2, and I'm just
squaring that.
So now here is a formula for the sum of
the first k perfect cubes.
It's k squared times k plus 1 squared over
4.
Right?
I just took this formula and squared it.
Again there's a slick geometric proof of
this fact.
So let's add 1 cubed plus 2 cubed plus 3
cubed plus 4 cubed pictorially, of course
the same arguments going to work in
general, if I add up the first k perfect
cubes.
Alright, well here's 1 cubed.
Alright.
Here's 2 cubed.
It's 8 dots, but I've arranged them in a,
in a cube.
Here's 3 cubed, all right?
It's 3 times 3 times 3, 27 dots, but I've
arranged them again in, in a cube shape.
And then 4 cubed would be 64 dots, but I
haven't drawn all the dots in, right?
So I want to count all of these dots, and
the trick is to rearrange these dots.
So that's how most of these sort of
pictorial arguments end up going.
So I'm going to rearrange these dots like,
like this.
So here I've made a square, and here's the
1 cube, I added just, just 1 dot.
Here's the 2 cube, but I've rearranged it
where 1 of the layers of the 2 by 2 by 2
cube is right here, and I've split the
other layer in half.
So if I take this vertical 2 by 1 piece
and this horizontal 1 by 2 piece, I can
rearrange them to make it 2 by 2 square
that I stack on top of that one, which
then gives me this again.
Here's the 3 by 3 by 3 cube, it's 3, 3 by
3 squares.
Here's the 4 by 4 by 4 cube.
It's 4 squares, but that top square is
been cut in half.
Right, so this piece here, and this piece
here, would combine to give me another 4
by 4 square, which I can stack on top of
these three to recover the 4 by 4 by 4
cube.
Now I could put five 5 by 5 squares
around, and then around that I could put
six 6 by 6 squares, but I'd cut one of
those squares in half just like this.
So the even layers I use square that's
been cut in half, and the odd layers, I
can just stack them down.
The, the point here is that I end up
producing a square whose side length is 1
plus 2 plus 3 plus 4.
You know, and of course this is being
general.
Alright, it would be plus k.
So this is 1 plus 2 plus 3 plus 4.
I've rearranged the sum of cubes into a
square, and that side length of that
square is the sum of the first k whole
numbers.
Good news now is that I've got a formula
for the sum of the first k whole numbers.
It's k times k plus 1 over 2.
So the resulting square has side length k
times k plus 1 over 2, which then means
that if I take the sum of the first K
cubes it's that number squared.
These kinds of formulas enable us to
perform calculations that would be
extraordinarily painful without access to
these formulas.
For instance, what if I wanted to add up
the first 100 perfect cubes, right.
That's the same thing as computing the
sum, n goes from 1 to 100, of n cubed.
Now, because I've got a formula for this,
because this is, the same as, just the sum
of The first hundred whole numbers
squared, right?
I know how to compute this, right?
What's the sum of the first hundred whole
numbers, well that's 100 times 101 over 2,
and I just have to square that.
What's a 100 times 101 over 2?
What's the sum of the first hundred
numbers?
It's 5,050 and I just have to square this.
And if I square this I get 2, 5, 5, 0, 2,
5, 0, 0.
All right?
So I get about 25 million, and this makes
possible, you know, this calculation that
would have been horrible tedious, right?
But because I've got access to this
formula, I can perform really amazing
numeric feats, right?
This would be horrible to calculate, but
with this formula, I can just write down
the answer.
If you like these kinds of formulas,
here's a challenge.
Here's the challenge.
What's the sum of fourth powers.
The answer's going to be some polynomial
in K, right, but you've gotta figure out
which polynomial that is.
One trick would be to plug in values.
Say K equals one, K equals two, K equals
three.
And then try to find the polynomial that
passes through those points.
Now that's your challenge.
And there's a long tradition here.
Bernoulli for instance computed the sum of
the first thousand tenths powers, and he
did it 300 years ago.
People have been thinking about these
kinds of formulas for a really long time.