A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Can anyone please explain The Binomial Theorem and Pascal’s Triangle to me? I will medal?!
anonymous
 one year ago
Can anyone please explain The Binomial Theorem and Pascal’s Triangle to me? I will medal?!

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Practice problems would be great too because I have no idea how to do it

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4First of all, recall that a "binomial" is a polynomial with "two" terms. so the theorem must have something to do with "two terms".

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Indeed it does, the binomial theorem gives a nice expression for repeated multiplication of a binomial with itself: \[(x+y)(x+y)(x+y)\cdots\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Without beating around the bush, here is the theorem in all its glory : \[\large (x+y)^n = \sum\limits_{k=0}^n \color{red}{\binom{n}{k}}x^ky^{nk}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'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").

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0:O Stop! The formula was NOT what I was expecting...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But how does it relate to the binomial theorem

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\(\large \color{red}{\binom{n}{k}}\) is called "binomial coefficient", pascal's triangle helps you in finding this coefficeint.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let’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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0uhhuh I'm processing this slowly, but I feel like I'm getting it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now take a look at the pascals triangle. (just search it up online)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0compare the coefficients, and powers with the pascals triangle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01 a + b a^2 + 2ab + b^2 a^3 + 3a^2b + 3ab^2 + b^3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, I understand that.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can any of you guys give me like a practice question to help maybe?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The row represents the coefficients

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hold on ffggff i just need to say one more thing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Take 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so that is the relationship between the two Here is a practice problem Expand (x + 2)^5 using the Binomial Theorem and Pascal’s triangle.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Side 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, I'm lost lol. So I drew the little triangle thingy and got 1, 5, 10, 10, 5, 1. But now what?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Like I can do the triangle but I don't know what to do afterwards?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So then \[1x^5 2^0 + 5x^4 2^1 + 10x^3 2^2 + 10x^2 2^3 + 5x^12^4 + 1x^02^5 =\] \[x^5 + 5 • 2x^4 + 10 • 4x^3 + 10 • 8x^2 + 5 • 16x + 32 =\] \[x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0:O Whoa! Okay! How did you add the 6th row of the triangle to the equation?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I mean, how did you get those exponents?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Remember 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok :) Thankyou, I get it now!!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So basically they are coming from (a+b)^5 in a way?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@zzr0ck3r I feel so bad! Lol you have been typing forever :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how 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?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1A 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 \((a+b)^7=a^7+7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6+b^7\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you see what I mean kinda?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1I did not realize it was going to be so long once I started...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@zzr0ck3r Thankyou :), I might just use your way from now on. It's simple yet very effective!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well yes @zzr0ck3r but ffggfgf wants to know the relationship between binomial theorem and pascals triangle.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1this actually shows it:)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1I can generally do these things and write them down faster than someone can put it in a calculator and write it down.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i use a different method than these, it is quite fast also.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1But anyway, it was meant to be a little extra addon to what you all were saying...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i would write it, but i don't want to spend another hour lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks to everyone who helped!! I really understand how to do it now!! :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4I think that works because of combinations, which term we pick in each binomial : dw:1436843111371:dw

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0one day see if you can find a nice proof of why pascal's triangle works for these questions, and also why the construction of the triangle give the entries \(\binom{n}{k}\) someday later on i mean

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks misty1212 :) Lol once I get to Ap cal I will!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then what do i do ganeshie?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4You will know what to do once you're faced with an example problem. Let me see if I can cookup an interesting problem :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Kinda scared now hahah :) But I will try

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Find the coefficient of \(x^6\) when you expand out below product : \[(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)\] (there are 7 terms in the product)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4nope, but how did you guess 1 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01 9 36 84 126 126 84 36 9 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I did the (a+b)^6 using the triangle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the first one would be one, as the first one would be x^7.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0remember what i said the exponents on the x part decrease from n. And btw, @ganeshie8 typed the expanded version of (x+1)^7

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohhh whoops But yes, go on.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Yes, maybe forget about the theorem/pascal's triangle and try working it pretending we never heard of these

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0tru tru, lets forget lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.0@misty1212 you wanted a proof for why the construction of pascal triangles has combinatoric solutions?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)\] Notice that there are \(7\) factors here and since we want the coefficient of \(x^6\), we must pick "x" from "6" factors and pick "1" from one factor. We can pick "1" from any of the 7 factors in exactly 7 ways. so the coefficeint of \(x^6\) in above product is simply \(7\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4As you can see, the "coefficient" is really equivalent to "number of ways of choosing a factor"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so would it be 1 7 21 35 35 21 7 1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4If you use pascal's triangle, yes.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Let me ask you a side question : Suppose you went for shopping and liked 7 different shirts, but you can only buy 6 shirts. how many total ways are there to choose 6 shirts from 7 shirts ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Let me change the question slightly : There are 7 shirts and you can buy only 1 shirt. How many ways are there to choose 1 shirt from 7 shirts ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.07? I don't really understand the question. Can u just explain it to me please?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Correct! I will explain after the drill in the end. lets try another related question : There are 7 shirts and you can buy only 1 shirt. How many ways are there to "reject" 6 shirts from 7 shirts ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Yes you do see that "choosing 1 shirt from 7 shirts" is same as "rejecting 6 shirts from 7 shirts"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0At first I didn't, but now I do.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4"choosing 1 girlfriend from available 7 friends" is same as "rejecting 6 girls from the available 7 friends"

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4sry for bad analogy but you get the point

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lol yeah, I get it :D

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4so what has that anything to do with finding the coefficient of \(x^6\) in the below product : \[(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)\] ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, okay so I'm assuming it would be 7

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes! Omg! Ok, I see. Visuals really do help!!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Maybe writing the product pictorially helps in seeing whats going on : dw:1436845412826:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4I feel that is enough for one session, just open a new thread if you still have any question :) good luck!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for all your help<3
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.