Question

Can anyone please explain The Binomial Theorem and Pascal’s Triangle to me? I will medal?!

Answer 1

Practice problems would be great too because I have no idea how to do it

Answer 2

First of all, recall that a "binomial" is a polynomial with "two" terms. so the theorem must have something to do with "two terms".

Answer 3

Ok:)

Answer 4

Indeed it does, the binomial theorem gives a nice expression for repeated multiplication of a binomial with itself:
\[(x+y)(x+y)(x+y)\cdots\]

Answer 5

I see, I see

Answer 6

Without beating around the bush, here is the theorem in all its glory :
\[\large (x+y)^n = \sum\limits_{k=0}^n \color{red}{\binom{n}{k}}x^ky^{n-k}\]

Answer 7

I'm just gonna go ahead and do pascals triangle.

In very simple words, its a number triangle.

To build the triangle, start with "1" at the top, then continue placing numbers below it in a triangular pattern.

Each number is the two numbers above it added together (except for the edges, which are all "1").

Answer 8

:O Stop! The formula was NOT what I was expecting...

Answer 9

But how does it relate to the binomial theorem

Answer 10

\(\large \color{red}{\binom{n}{k}}\) is called "binomial coefficient", pascal's triangle helps you in finding this coefficeint.

Answer 11

Let’s explore the first couple of powers of the binomial (a + b).

(a + b)0 = 1 *Remember, anything to the zero power equals 1.
(a + b)1 = a + b
(a + b)2 = (a + b)(a + b) = a^2 + 2ab + b^2
(a + b)3 = (a + b)(a2 + 2ab + b2) = a^3 + 3a^2b + 3ab^2 + b^3

Answer 12

uh-huh I'm processing this slowly, but I feel like I'm getting it

Answer 13

Now take a look at the pascals triangle.
(just search it up online)

Answer 14

compare the coefficients, and powers with the pascals triangle

Answer 15

1
a + b
a^2 + 2ab + b^2
a^3 + 3a^2b + 3ab^2 + b^3

Answer 16

and
1
11
121
1331

Answer 17

Yes, I understand that.

Answer 18

Can any of you guys give me like a practice question to help maybe?

Answer 19

The row represents the coefficients

Answer 20

Hold on ffggff i just need to say one more thing

Answer 21

kk:)

Answer 22

Take a look at the exponents for (a+b)^3 The powers of a decrease from 3 to 0, and the powers of b increase from 0 to 3.

Answer 23

so that is the relationship between the two

Here is a practice problem

Expand (x + 2)^5 using the Binomial Theorem and Pascal’s triangle.

Answer 24

Side note, is that the fifth row of pascals triangle would represent (a+b)^4, and the third would be (a+b) ^2. The reason is that the first row, represents (a+b)^0

Answer 25

Ok, I'm lost lol. So I drew the little triangle thingy and got 1, 5, 10, 10, 5, 1. But now what?

Answer 26

Like I can do the triangle but I don't know what to do afterwards?

Answer 27

The sixth row of Pascal’s triangle would be 1, 5, 10, 10, 5, 1 and the exponents would start and end with 5. And that is what you got

Answer 28

Yup

Answer 29

So then
\[1x^5 2^0 + 5x^4 2^1 + 10x^3 2^2 + 10x^2 2^3 + 5x^12^4 + 1x^02^5 =\]
\[x^5 + 5 • 2x^4 + 10 • 4x^3 + 10 • 8x^2 + 5 • 16x + 32 =\]
\[x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32\]

Answer 30

:O Whoa! Okay! How did you add the 6th row of the triangle to the equation?

Answer 31

I mean, how did you get those exponents?

Answer 32

Remember this? Take a look at the exponents for (a+b)^3 The powers of a decrease from 3 to 0, and the powers of b increase from 0 to 3.

SO basically, as it goes, the powers of x decrease from the n to 0 (where n is the exponent, or (x+y)^n) And the powers of the other one (in this case it is 2) increases from 0 to n.

Answer 33

Ok :) Thank-you, I get it now!!!!

Answer 34

NP!!! :)

Answer 35

So basically they are coming from (a+b)^5 in a way?

Answer 36

@zzr0ck3r I feel so bad! Lol you have been typing forever :)

Answer 37

well, yes,

So to finalize it let the equation be (a+b)^n

So a's exponents would go from n to 0 and b's would increase from 0 to n.

Answer 38

how do you what to do to break it up into 5? Like how the first one is 1x^5 2^0 and the second one is 5x^42^1?

Answer 39

A neat trick you can use instead of the triangle.
Example:

\((a+b)^4\)

We know that if will look something like this

\(\text{_}a^4+\text{_}a^3b+\text{_}a^2b^2+\text{_}ab^3+\text{_}b^4\)

Now the question is, what do we put in the _ places

Well we know the first coefficient will be \(1\).

For the second _ we multiply the power of \(a\) in the first term, by the coefficient in the first term and then divide that all by \(1\) (we use one because the previous two steps were in the first term). So \((4*1)/1=4\) and that is our second coefficient.

so we have \(a^4+4a^3b+\text{_}a^2b^2+\text{_}ab^3+\text{_}b^4\)

For the third coefficient, we multiply the power of \(a\) in the second term by the coefficient in the second term and divide by \(2\) (we are doing stuff in the second term now) and we get \(3*4/2 = 6\) and that is our third coefficient.



Similarly our fourth coefficient is \(2*6/3=4\) and our last coefficient is \(1*4/4=1\)

So we have \(a^4+4a^3b+6a^2b^2+4ab^3+b^4\). Also I am sure someone pointed this out above, but once we find the first half of the coefficients, we are done (notice the pattern).

Ok this all seems very hard and long, but that was just because I was doing lots of steps.

If wanted to do \((a+b)^7\)

First we quickly write out

\(a^7+a^6b+a^5b^2+a^4b^3+a^3b^4+a^2b^5+ab^6+b^7\)

and note the leading coefficient is \(1\).

then we do

\(7*1/1=7\\

6*7/2=21\\

5*21/3=5*7=35\)

and we are done because we have gone half way

so we have \((a+b)^7=a^7+7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6+b^7\)

Answer 40

do you see what I mean kinda?

Answer 41

I did not realize it was going to be so long once I started...

Answer 42

@zzr0ck3r Thank-you :), I might just use your way from now on. It's simple yet very effective!

Answer 43

well yes @zzr0ck3r but ffggfgf wants to know the relationship between binomial theorem and pascals triangle.

Answer 44

this actually shows it:)

Answer 45

It's a good method.

Answer 46

I can generally do these things and write them down faster than someone can put it in a calculator and write it down.

Answer 47

i use a different method than these, it is quite fast also.

Answer 48

But anyway, it was meant to be a little extra add-on to what you all were saying...

Answer 49

i would write it, but i don't want to spend another hour lol

Answer 50

Thanks to everyone who helped!! I really understand how to do it now!! :)

Answer 51

I think that works because of combinations, which term we pick in each binomial :