anonymous
  • anonymous
Can anyone please explain The Binomial Theorem and Pascal’s Triangle to me? I will medal?!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Practice problems would be great too because I have no idea how to do it
ganeshie8
  • ganeshie8
First of all, recall that a "binomial" is a polynomial with "two" terms. so the theorem must have something to do with "two terms".
anonymous
  • anonymous
Ok:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ganeshie8
  • ganeshie8
Indeed it does, the binomial theorem gives a nice expression for repeated multiplication of a binomial with itself: \[(x+y)(x+y)(x+y)\cdots\]
anonymous
  • anonymous
I see, I see
ganeshie8
  • ganeshie8
Without beating around the bush, here is the theorem in all its glory : \[\large (x+y)^n = \sum\limits_{k=0}^n \color{red}{\binom{n}{k}}x^ky^{n-k}\]
anonymous
  • anonymous
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").
anonymous
  • anonymous
:O Stop! The formula was NOT what I was expecting...
anonymous
  • anonymous
But how does it relate to the binomial theorem
ganeshie8
  • ganeshie8
\(\large \color{red}{\binom{n}{k}}\) is called "binomial coefficient", pascal's triangle helps you in finding this coefficeint.
anonymous
  • anonymous
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
anonymous
  • anonymous
uh-huh I'm processing this slowly, but I feel like I'm getting it
anonymous
  • anonymous
Now take a look at the pascals triangle. (just search it up online)
anonymous
  • anonymous
compare the coefficients, and powers with the pascals triangle
anonymous
  • anonymous
1 a + b a^2 + 2ab + b^2 a^3 + 3a^2b + 3ab^2 + b^3
anonymous
  • anonymous
and 1 11 121 1331
anonymous
  • anonymous
Yes, I understand that.
anonymous
  • anonymous
Can any of you guys give me like a practice question to help maybe?
anonymous
  • anonymous
The row represents the coefficients
anonymous
  • anonymous
Hold on ffggff i just need to say one more thing
anonymous
  • anonymous
kk:)
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
Ok, I'm lost lol. So I drew the little triangle thingy and got 1, 5, 10, 10, 5, 1. But now what?
anonymous
  • anonymous
Like I can do the triangle but I don't know what to do afterwards?
anonymous
  • anonymous
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
anonymous
  • anonymous
Yup
anonymous
  • anonymous
So then \[1x^5 2^0 + 5x^4 2^1 + 10x^3 2^2 + 10x^2 2^3 + 5x^12^4 + 1x^02^5 =\] \[x^5 + 5 • 2x^4 + 10 • 4x^3 + 10 • 8x^2 + 5 • 16x + 32 =\] \[x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32\]
anonymous
  • anonymous
:O Whoa! Okay! How did you add the 6th row of the triangle to the equation?
anonymous
  • anonymous
I mean, how did you get those exponents?
anonymous
  • anonymous
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.
anonymous
  • anonymous
Ok :) Thank-you, I get it now!!!!
anonymous
  • anonymous
NP!!! :)
anonymous
  • anonymous
So basically they are coming from (a+b)^5 in a way?
anonymous
  • anonymous
@zzr0ck3r I feel so bad! Lol you have been typing forever :)
anonymous
  • anonymous
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.
anonymous
  • anonymous
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?
zzr0ck3r
  • zzr0ck3r
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 \((a+b)^7=a^7+7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6+b^7\)
anonymous
  • anonymous
do you see what I mean kinda?
zzr0ck3r
  • zzr0ck3r
I did not realize it was going to be so long once I started...
anonymous
  • anonymous
@zzr0ck3r Thank-you :), I might just use your way from now on. It's simple yet very effective!
anonymous
  • anonymous
well yes @zzr0ck3r but ffggfgf wants to know the relationship between binomial theorem and pascals triangle.
zzr0ck3r
  • zzr0ck3r
this actually shows it:)
anonymous
  • anonymous
It's a good method.
zzr0ck3r
  • zzr0ck3r
I can generally do these things and write them down faster than someone can put it in a calculator and write it down.
anonymous
  • anonymous
i use a different method than these, it is quite fast also.
zzr0ck3r
  • zzr0ck3r
But anyway, it was meant to be a little extra add-on to what you all were saying...
anonymous
  • anonymous
i would write it, but i don't want to spend another hour lol
anonymous
  • anonymous
Thanks to everyone who helped!! I really understand how to do it now!! :)
ganeshie8
  • ganeshie8
I think that works because of combinations, which term we pick in each binomial : |dw:1436843111371:dw|
misty1212
  • misty1212
one 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
  • anonymous
Thanks misty1212 :) Lol once I get to Ap cal I will!
anonymous
  • anonymous
Then what do i do ganeshie?
ganeshie8
  • ganeshie8
You will know what to do once you're faced with an example problem. Let me see if I can cookup an interesting problem :)
anonymous
  • anonymous
Kinda scared now hahah :) But I will try
ganeshie8
  • ganeshie8
Find 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)
anonymous
  • anonymous
Ok, hold on please
anonymous
  • anonymous
Would it be 1?
ganeshie8
  • ganeshie8
nope, but how did you guess 1 ?
anonymous
  • anonymous
1 9 36 84 126 126 84 36 9 1
anonymous
  • anonymous
I did the (a+b)^6 using the triangle
anonymous
  • anonymous
the first one would be one, as the first one would be x^7.
anonymous
  • anonymous
remember what i said the exponents on the x part decrease from n. And btw, @ganeshie8 typed the expanded version of (x+1)^7
anonymous
  • anonymous
ohhh whoops But yes, go on.
ganeshie8
  • ganeshie8
Yes, maybe forget about the theorem/pascal's triangle and try working it pretending we never heard of these
anonymous
  • anonymous
tru tru, lets forget lol
dan815
  • dan815
@misty1212 you wanted a proof for why the construction of pascal triangles has combinatoric solutions?
ganeshie8
  • ganeshie8
\[(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)\] 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 \(7\)
anonymous
  • anonymous
Ok!
ganeshie8
  • ganeshie8
As you can see, the "coefficient" is really equivalent to "number of ways of choosing a factor"
anonymous
  • anonymous
so would it be 1 7 21 35 35 21 7 1
anonymous
  • anonymous
Yup!
ganeshie8
  • ganeshie8
If you use pascal's triangle, yes.
anonymous
  • anonymous
Gotcha
ganeshie8
  • ganeshie8
Let 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 ?
dan815
  • dan815
|dw:1436844308425:dw|
anonymous
  • anonymous
6
ganeshie8
  • ganeshie8
think again
anonymous
  • anonymous
6x7 basically
ganeshie8
  • ganeshie8
Let 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
  • anonymous
7? I don't really understand the question. Can u just explain it to me please?
ganeshie8
  • ganeshie8
Correct! 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 ?
anonymous
  • anonymous
7?
ganeshie8
  • ganeshie8
Yes you do see that "choosing 1 shirt from 7 shirts" is same as "rejecting 6 shirts from 7 shirts"
anonymous
  • anonymous
yup. I do see.
anonymous
  • anonymous
At first I didn't, but now I do.
ganeshie8
  • ganeshie8
"choosing 1 girlfriend from available 7 friends" is same as "rejecting 6 girls from the available 7 friends"
ganeshie8
  • ganeshie8
sry for bad analogy but you get the point
anonymous
  • anonymous
Lol yeah, I get it :D
ganeshie8
  • ganeshie8
so 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
  • anonymous
oh, okay so I'm assuming it would be 7
anonymous
  • anonymous
Yes! Omg! Ok, I see. Visuals really do help!!
ganeshie8
  • ganeshie8
Maybe writing the product pictorially helps in seeing whats going on : |dw:1436845412826:dw|
ganeshie8
  • ganeshie8
I feel that is enough for one session, just open a new thread if you still have any question :) good luck!
anonymous
  • anonymous
Thanks for all your help<3

Looking for something else?

Not the answer you are looking for? Search for more explanations.