[MUSIC]. 
Let's find the sum of the 4th powers. 
In particular, I want to calculate the 
sum, n goes from 1 to k, of n to the 4th. 
Playing with this analogy, I could think 
about this problem the same way that I'd 
approach an integration problem. 
I should try to antidifference n to the 
4th. 
I should try to find a list of numbers, 
whose successive differences are n to the 
fourth, just try to find a list of 
numbers whose differences are n to the 
4th. 
Let's just try to find some lists of 
numbers and their differences. 
So as a first example let's take a look 
at say n to the 5th, and let's look at 
the 5th power. 
So 0 the 5th is 0, 1 to the 5th is 1, 2 
to the 5th is 32, 3 to the 5th is 243, 4 
to the 5th is the same as 2 to the tenth 
that's 1024. 
Five to the 5th is 3125 and so on. 
So that's just lists of of 5th powers. 
Now, let's try to find the differences 
between subsequent numbers in this list. 
So the difference between 1 and 0 is 1. 
The difference between 32 and 1 is 31. 
The difference between 243 and 32 is 211. 
The difference between 1,024 and 243 is 
781. 
And the difference between 1,024 and 
3,125 is 2101 and so forth. 
Now it's maybe not so helpful just to see 
specific numbers. 
Let's try to write down a formula for 
this. 
So I'll write the differences, between 
the 5th powers. 
Well that's n to the 5th minus the 
previous 5th power, so minus n minus 1 to 
the 4th. 
And then I could expand this out. 
I could take n to 5th minus this thing 
expanded and I'd get 5 n to the 4th is 
the n to the 5th will cancel minus 10 n 
cubed plus 10 n squared minus 5 n plus 1. 
Right, and the plus 1 comes from 
subtracting a minus 1 to the 5th. 
Okay, so that formula there gives these 
green numbers, it's the differences in 
the 5th powers. 
I'm going to play the same kind of game 
for a different list of numbers. 
Let's try to find the differences in the 
list of say 4th powers, well that will be 
the 4th power of n, minus the fourth 
power of the pervious number of n minus 
1. 
And again I can expand this out and n to 
the 4th minus, there's an n to the fourth 
that cancels. 
So the highest power that survives is an 
n cubed term with a coefficient of 4 
minus 6 n squared plus 4 n minus 1. 
So this formula tells me the differences 
between subsequent 4th powers. 
Now I want to combine those, to get 
somehting with a difference of n to the 
4th. 
Okay, but how am I going to do that? 
Well, here's something I could do to make 
it a little bit easier to see what's 
going on. 
Instead of looking at differences between 
5th powers, I can look at the differences 
between 1 5th of 5th power. 
So I, that is the effect of dividing all 
these numbers by five. 
And since all these coefficients except 
for the last one are multiples of five, 
that'll make that formula look a little 
bit nicer. 
And I could do the same thing with with n 
to the 4th. 
Instead of looking at 4th powers, I could 
look at 1 quarter of fourth powers, and 
I'd have the effect of dividing all of 
these coefficients by by four. 
Let me just write down you know, what, 
what we've got here, kind of summarize 
the resulting formulas. 
So, the differences between n to the 5th 
over 5, well, according to this that will 
be n to the 4th minus 2 n cubed plus 2 n 
squared minus n plus a 5th. 
And I get a similar kind of formula for 
looking at differences of n to the 4th 
over 4. 
It's this divided by 4 and that will give 
me an n, cubed minus 3 halves and squared 
plus n minus a quarter. 
Well, let's just try to combine those 
two. 
But how, exactly, do I want to combine 
them? 
Well, the deal is that I've got an n to 
he 4th here, and that's really what I 
want at the end of this process. 
I want something with the differences n 
to the 4th, because I'm trying to 
antidifference n to the 4th. 
And I've got a minus 2n cubed. 
And I've just got an n cubed here. 
So if I took this and added two copies of 
this, I'd be in pretty good shape. 
So what would I get in that case? 
I'll be looking at the differences of n 
to the 5th over 5 plus two copies of n to 
the 4th over 4. 
And that would give me an n to the 4th. 
The n cubed term would go away. 
I'd get a minus n squared, because I've 
got a 2n squared minus two copies of 
three halves n squared. 
And then I get a plus n, because I've got 
minus n plus two copies of n. 
And then a 5th minus 2 copies of a 4th 
will give me minus 3 tenths. 
I could try to get rid of that n squared 
term by adding up on the differences of n 
cubed. 
Why does that work? 
Well, if you calculate the differences 
for the list of numbers n cubed over 3. 
you end up getting, n squared minus n, 
plus a third. 
So, if I take this and add this, that'll 
get rid of this n squared term. 
Right? 
So I'm going to look at the differences 
of n to the 5th over 5 plus two copies of 
n to the 4th over 4 plus n cubed over 3, 
and I'm left with an n to the fourth. 
No more n squared anymore. 
The plus n minus n also cancels, and then 
I've got minus three tenths plus a third, 
and that ends up being plus 1 30th. 
Now I can get rid of the 1 over 30th 
term. 
Now how do I do that? 
Well, look at the differences of the list 
of numbers n over 30, and that's just 1 
30th, right. 
This list of numbers, each number in that 
list differs by 1 30th compared to the 
previous number in the list. 
So if I take this and subtract n over 30, 
then the differences in that list are 
exactly n to the 4th, right? 
By which I mean that I take d of n to the 
5th over 5 plus 2 n to the 4th over 4 
plus n to the 3rd over 3 minus n over 30. 
And then I got n to the 4th plus a 30th 
minus a 30th and what I'm left with is 
just n to the 4th. 
So in light of all of this, What do I 
know right know? 
Taking differences between a list of 
numbers, and this accumulation function, 
adding up the first k numbers on the 
list. 
Those are inverse operations, right? 
This is the discreet version of the 
fundamental theorem of calculus. 
In this particular case, the differences 
between these numbers give you the 4th 
power. 
So if I sum the 4th powers I get this, at 
least up to some constant. 
Now I just have to worry about that 
constant. 
But then constant is zero, we can check 
it. 
Here's a sum for n equals 1 just to 1 of 
n to the 4th. 
And I can just plug in any value of K 
that I like, since this constant C 
doesn't depend upon K. 
I can figure out what that constant is by 
looking at when K equals 1. 
So, this is just 1. 
But on the other side, I've got what I 
get when I plug in k equals 1. 
Which is, 1 5th plus 2 times the 4th, 
plus, you know, 1 to the 3rd over 3. 
So plus a 3rd minus a 30th, plus that 
constant c that doesn't depend upon k at 
all. 
But a 5th plus a half plus a third minus 
a 30th is 1. 
So I've got 1 equals 1 plus C. 
So that means C equals 0. 
And that tells me what the formula then, 
for the sum of the 4fourh powers is. 
It's just this, and that constant is just 
0. 
This whole process is really analogous to 
the fundamental theorem of calculus and 
how we use it, right. 
In this example I'm antidifferencing n to 
the 4th in order to sum n to the 4th. 
In the same way that if I were to 
antidifferentiate x to the fourth, that 
would help me to integrate x to the 4th, 
right. 
This analogy runs really deep.