We have these three terms here in the same columns.

So as you see when you compute the sum of them, this is essentially the same

as computing, so this 4 comes as the sum of 3 and 1.

This 6 comes as a sum of 3 and 3.

And this 4 comes as the sum of 1 and 3.

This is exactly the pattern in our Pascal's triangle.

And this is true in general, not only for n equal to 4 and 3 of course.

So this way of proving the binomial theorem by induction, okay?

Now let me show you also some slightly more complicated example,

just to show you that binomial theorem allows us to compute the coefficients

of all the monomials, not only for the case when we had just a + b.

Where instead we might want to compute the 4th

power of not just a+b but something like 2a-b.

In this case, what remains to be done is just to consider 2a-b as our new a and b.

So we consider 2a-b as the sum of 2a and -b.

And then for this, for

the sum of these two guys we are applying the binomial theorem.

This gives us 2a to the 4th,

then 4 times 2a cubed times -b times our second friend and

6 times 2a squared times -b squared and so on.

And then you just need to multiply these binomial

coefficients with powers of 2 coming from 2a, okay?

And this finally gives you the following expression.

So, in particular, if you have something like 2a- b, to the 7th,

for example, you don't need to actually multiply many, many, many terms.

You just need to get the seventh row of Pascal triangle,

and this will already give you all the coefficients.

All what remains to be done is probably to multiply them with coefficients like 2,

if instead of a+b, you have something like 2a+b, okay?

Now let's also show several interesting consequences of the binomial theorem.

For example, if we set a and b = 1, then what we get on the left,

so instead of a plus b to the n, what we get is 1 plus 1 to the n,

that is just 2 to the n, okay?

And this is what we get here.

On the other hand, a to the k and

b to the n minus k is always equal to 1, because a and b is equal to 1.

So on the right, what we have is the sum of all binomial coefficients.

So this gives us another proof of the fact that the sum of all

binomial coefficients is equal to 2 to the n, okay?

On the other hand, if we set a equal to 1,

and b equal to -1, and on the left we have 0, right?

Because we have 1 + -1 to the n.

On the other hand, on the right what we have is just an alternating sum, okay?

So this gives us another proof of the fact that the alternating sum is equal to 0.

Let me recall, you also, the combinatorial, this means that for

any subset, for any n greater than 0,

the number of odd size subsets is the same as the number of even size subsets.

And the combinatorial meaning of the previous equation is

the number of subsets of known element set is equal to the 2 the n.

Okay, finally let me show you one more sequence,

the one that we haven't seen before.

So if we set a =1 and b =2, then what we get on the left,

we have a+b which is 3, so 3 to the n.

And on the right, we have binomial coefficients and

each binomial coefficient n choose k is multiplied by 2 to the k, okay?

We have this nice, nice formula.

3 to the n is equal to n choose 0 + n choose 1 times 2 +

n choose 2 times 2 squared + and so on, n choose n times 2 to the n.

So this is an interestingly looking example but

it nicely you'd also have a combinatorial meaning.

So let me give you a combinatorial proof of this equation.

So what we have on the left here may be viewed as a number of ways of composing

a word of length n out of three symbols, we know them by x, y and z.

So 3 to the n is of course the number of such words because for

every of n positions we have three possibilities, x, y and z.

So now our goal is to express this number of word as this sum.

And this is what we're going to do now.

So let's compute it as follows.

So if n chose 0 is the set is the number of all words

of lengths n that consist of only of letter x, okay?

So n chose 0 is actually 1.

So when there is exactly one such word, so

it is just x, x, x, x, n times, right?

Okay, so we justified this.

Now, what we're going to compute next is a number of words

containing exactly n-1 letters x, okay?

So how to compute it and why it is equal to n choose 1 times 2?

Well, it is equal to this expression because of the following fact.

To compose such a work,

we first need to select exactly one position where x does not appear.

So we can do this in this many ways, n choose 1.

And then for this position we need to choose either y or z.

So we multiply by 2, okay?

So this is once again the number of words of things n that

contain exactly n-1 letters x.

Okay, let me find and then show you the next one, and the pattern will emerge.

So, this says as you might guessed already exactly the number

of words of length n that can contain exactly n-2 letters x, y, z.

Well, again,

because to form such word we first need to select two places where there is no x.

And for each of these two places, we need to select either y or z.

So first of all, there are n choose 2 choices of these two places, and then for

each of these two choices, we need to select either y or z.

So it is 2 times 2, okay?

And this gives us n choose 2 times 2 squared.

And in general, if we would like to compute the sum of

the number of words that contain exactly n- k letters x,

then we first select k positions where x does not appear.

And then for each of these k positions,

we select either y or z which gives us n choose k times 2 to the k