# Math Help - Pascals Triangle

1. ## Pascals Triangle

Hi, would some one please explain to me the equation in the beginning of this video and what it is solving and provide an example, and just explain what the pascal triangle is and what it accomplishes, because i don't get it >.<

please and thank you. I need this because part of my math summative is going to be on this and I'm supposed to learn this by myself!

2. ## Re: Pascals Triangle

Originally Posted by sakonpure6
Hi, would some one please explain to me the equation in the beginning of this video and what it is solving and provide an example, and just explain what the pascal triangle is and what it accomplishes, because i don't get it, please and thank you. I need this because part of my math summative is going to be on this and I'm supposed to learn this by myself!

Frankly I don't understand what you are confused about.

Do you understand that if $N\ge k,~\binom{N}{k}=\frac{N!}{k!(N-k)!}~?$

Example: $\binom{10}{4}=\frac{10!}{4!(6!)}$

Now all the video says that in the expansion of $(a+b)^N$ each term looks like $\binom{N}{k}a^kb^{N-k}$

3. ## Re: Pascals Triangle

Okay... and this binomial theorem is just a faster way of finding the coefficient of for example (x+3)^5 rather than multiplying it out?

4. ## Re: Pascals Triangle

Originally Posted by sakonpure6
Okay... and this binomial theorem is just a faster way of finding the coefficient of for example (x+3)^5 rather than multiplying it out?
Yes, the Binomial Theorem enables you to expand out binomials quickly without having to physically multiply out each term.

5. ## Re: Pascals Triangle

Originally Posted by sakonpure6
Okay... and this binomial theorem is just a faster way of finding the coefficient of for example $(x+3)^5$ rather than multiplying it out?
Suppose that we want to know the term that contains $x^3$ it would be $\binom{5}{3}(3)^2x^3$

6. ## Re: Pascals Triangle

Although it is possible to use Pascal's triangle to find the coefficients in the expansion of $(x+y)^n$ where $n\in\mathbb{N}_0$, this becomes impractical for large values of $n$ because of the need to write out all of the preceding rows. For this reason, let's look at another pattern in the expansion of $(x+y)^n$.

The first and last terms are $x^n$ and $y^n$ respectively. In this expansion, the coefficient of any term after the first can be generated as follows:

In the previous term, which we'll enumerate as the $i$th term, multiply the coefficient by the exponent of $x$ and then divide by $i$. I'll demonstrate this technique for the expansion of $(x+y)^6$.

The first term could be written as:

$1\cdot x^6y^0$

For the next term, we subtract 1 from the exponent of $x$, add 1 to the exponent of $y$ and find that the coefficient is $\frac{1\cdot6}{1}=6$ and so the next term is $6x^5y$.

And so the third term is $\frac{6\cdot5}{2}x^4y^2=15x^4y^2$, and then the fourth term is $\frac{15\cdot4}{3}x^3y^3=20x^3y^3$.

Now, by symmetry, where observation of the exponents tells you when you have reached the mid-point, that is, when the exponent on $x$ is less than or equal to the exponent on $y$, we then know the coefficients will on the right be the "mirror image" of those on the left, and we may now state:

$(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^2+15x^2y^4+6xy^5 +y^6$

As you can see, this method is more efficient than using Pascal's triangle, because as stated, there is no need to generate the coefficients for every line previous to the one you want. You do have to generate the first half of a line and then use symmetry to fill in the other half.

Now, as Plato pointed out, the most efficient way of finding these binomial coefficients uses factorial notation. The number $n!$ (read "n factorial) is defined as follows:

For any natural number $n$:

$n!=n(n-1)(n-2)\cdots3\cdot2\cdot1$

It can be proven by induction that the coefficient of the $i$th term ( $i$ = 0 to $n$) in the expansion of $(x+y)^n$ is:

$\frac{n!}{i!(n-i)!}\equiv{n \choose i}$ where $i\le n$.

This number, called a binomial coefficient, is also referred to as $n$ above $i$ or $n$ choose $i$, since it can be shown that this is the number of way to choose $i$ elements from a set of $n$ elements.

In this expansion, we will find:

i) There are $n+1$ terms.

ii) The first term is $x^n$, and the last term is $y^n$.

iii) The exponents of $x$ decreases by 1, while the exponent on $y$ increases by 1 in each succeeding term.

iv) The sum of the exponents on $x$ and $y$ is $n$ in every term.

v) The coefficient of $x^{n-r}y^r$ and $x^ry^{n-r}$ is ${n \choose r}$.

vi) The $i$th term is ${n \choose i-1}x^{n+1-i}y^{i-1}$.

These observations about the expansion of $(x+y)^n$ suggest the binomial theorem:

$(x+y)^n=\sum_{k=0}^n\left({n \choose k}x^{n-k}y^k \right)$

Here is another way of looking at the binomial theorem. Consider the following expansion:

$(x+y)^3=(x+y)(x+y)(x+y)=$

$xxx+xxy+xyx+xyy+yxx+yxy+yyx+yyy$

You obtain each product shown in the expansion by multiplying 3 variables, one from each of the binomial factors $(x+y)$. The term $yxx$ for example is the result of choosing $y$ from the first binomial factor, $x$ from the second and $x$ from the third. Do you see that the products $xxy$, $xyx$ and $yxx$ are those that result when you select $y$ from one binomial factor and $x$ from the other two?

Can you see how this implies $\sum_{k=0}^n{n \choose k}=2^n$? This of each variable as a binary digit and notice we are simply counting from 0 to $2^n-1$.

So, by the commutative property of multiplication, we may combine like terms to get:

$(x+y)^3=x^3+3x^2y+3xy^2+y^3$

and so we may observe that 3 is the number of ways to select either 1 $x$ or 1 $y$ from the three binomial factors.

We can think of each binomial factor as the flip of a coin, where there are two possible outcomes. The binomial coefficient just shows how many different ways there are to obtain $r$ of one outcome and $n-r$ of the other.

7. ## Re: Pascals Triangle

omg I thank you guys from the bottom of my heart! I really appreciate this!