Sal expands (3y^2+6x^3)^5 using the binomial theorem and Pascal's triangle. Created by Sal Khan.
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- This problem is a bit strange to me. Sal says that "We've seen this type problem multiple times before." I haven't. I must have missed several videos along the way. Can someone point me in the right direction? Thank's very much.(12 votes)
- If you are looking for videos relating to the Binomial Theorem and Pascal's Triangle, try these videos:
If you are looking for videos involving combinatorics, look here:
- What does Sal mean by 5 choose 1?(6 votes)
- 5C1 or 5 choose 1 refers to how many combinations are possible from 5 items, taken 1 at a time. That means that 5 choose 1 is 5.
This topic is related to probability (permutations and combinations). You might want to check out that topic if you don't understand.
(My Source: http://www.statisticshowto.com/5-choose-3-5c3-figuring-combinations/)(11 votes)
- Wow. This makes absolutely zero sense whatsoever.(9 votes)
- Its just a specific example of the previous binomial theorem where a and b get a little more complicated. Rather than figure out ALL the terms, he decided to hone in on just one of the terms. Sometimes in complicated equations, you only care about 1 or two terms.(2 votes)
- how do we solve this type of problem when there is only variables and no numbers?(6 votes)
- The only way I can think of is (a+b)^n where you would generalise all of the possible powers to do it in, but thats about it, in all other cases you need to use numbers(3 votes)
- how do you know if you have to find the coefficients of x6y6(5 votes)
- at5:37, what are the exclamation marks (factorials)? do they symbolize a number?(1 vote)
- If n is a positive integer, then n! means "n factorial", which is defined as the product of the positive integers from 1 to n inclusive (for example, 4! = 1*2*3*4 = 24). Furthermore, 0! is defined as 1.
The exclamation mark is itself not a number, but it can be thought of as a function (or unary operation).
Have a blessed, wonderful day!(5 votes)
- how do you do it when the equation is (a-b)^7? When the sign is negative, is there a different way of doing it? If there is a new way, why is that?(2 votes)
- The only difference is the 6x^3 in the brackets would be replaced with the (-b), and so the -1 has the power applied to it too. Odd powered brackets would therefore give negative terms and even powered brackets would gve a positive term.(2 votes)
- The coefficient of x^2 in the expansion of (1+x/5)^n is 3/5, (i) Find the value of n
(ii) With this value of n, find the term independent of x in the expansion (1+x/5)^n(2-3/x)^2(2 votes)
- sounds like we want to use pascal's triangle and keep track of the x^2 term. We can skip n=0 and 1, so next is the third row of pascal's triangle.
1 2 1 for n = 2
the x^2 term is the rightmost one here so we'll get 1 times the first term to the 0 power times the second term squared or 1*1^0*(x/5)^2 = x^2/25 so not here.
1 3 3 1 for n = 3
Squared term is second from the right, so we get 3*1^1*(x/5)^2 = 3x^2/25 so not here
1 4 6 4 1 for n = 4
Squared term is the third from the right so we get 6*1^2*(x/5)^2 = 6x^2/25
1 5 10 10 5 1 for n = 5
Squared term is fourth from the right so 10*1^3*(x/5)^2 = 10x^2/25 = 2x^2/5 getting closer.
1 6 15 20 15 6 1 for n=6
Fifth from the right here so 15*1^4*(x/5)^2 = 15x^2/25 = 3x^2/5 There we are.
So n has to equal 6 Now you can use this line of the triangle to find the term where the x part is brought to the 0 power.
If you're unfamiliar with pascal's triangle this probably doesn't make sense, and I'd be happy to explain it.(2 votes)
- What is combinatorics?(1 vote)
- Combinatorics is the branch of math about counting things. Combinatorial problems are things like 'How many ways can you place n-many items into k-many boxes, given that each box must contain at least 3 items? What if some of the items are identical?'(3 votes)
- can someone please tell or direct me to the proof/derivation of the binomial theorem.
it would be much appreciated(2 votes)
Voiceover:So we've got 3 Y squared plus 6 X to the third and we're raising this whole to the fifth power and we could clearly use a binomial theorem or pascal's triangle in order to find the expansion of that. But what I want to do is really as an exercise is to try to hone in on just one of the terms and in particular I want to hone in on the term that has some coefficient times X to the sixth, Y to the sixth. So in this expansion some term is going to have X to the sixth, Y to sixth and I want to figure out what the coefficient on that term is and I encourage you to pause this video and try to figure it out on your own. So I'm assuming you've had a go at it and you might have at first found this to be a little bit confusing. I'm only raising it to the fifth power, how do I get X to the sixth, Y to the sixth? But then when you look at the actual terms of the binomial it starts to jump out at you. Okay, I have a Y squared term, I have an X to the third term, so when I raise these to powers I'm going to get, I could have powers higher than the fifth power. But to actually think about which of these terms has the X to the sixth, Y to the sixth, let's just look at the pattern in, in I guess the actual expansion without even thinking about its coefficients. And we've seen this multiple times before where you could take your first term in your binomial and you could start it off it's going to start of at a, at the power we're taking the whole binomial to and then in each term it's going to have a lower and lower power. So let me actually just copy and paste this. So this is going to be, so copy and so that's first term, second term, let me make sure I have enough space here. Second term, third term, fourth term, fourth term, fifth term, and sixth term it's going to have 6 terms to it, you always have one more term than the exponent. And then, actually before I throw the exponents on it, let's focus on the second term. So the second term, actually I'll write it like this. So the second term's times 6 X to the third, let me copy and paste that, whoops. So let me copy and paste that. So we're going to put that there. That there. And that there. And that there. And then over to off your screen. I wrote it over there. We'll see if we have to go there. And then let's put the exponents. So this exponent, this is going to be the fifth power, fourth power, third power, second power, first power and zeroeth power. And for the blue expression, for 6 X to the third, this is going to be the zeroeth power, first power, first power, second power, third power, fourth power, and then we're going to have the fifth power right over here. Then and, of course, they're each going to have coefficients in front of them. They're each going to have coefficients in front of them. So there's going to be a coefficient in front of this one, in front of this one, in front of this one and then we add them all together. But which of these terms is the one that we're talking about. Has X to the sixth, Y to the sixth. So here we have X, if we take Y squared to the fourth it's going to be Y to the eighth, so that's not it. if we go here we have Y squared to the third power, that's Y to the sixth and here you have X to the third squared, that X to the sixth. So what we really want to think about is what is the coefficient, what is the coefficient in front of this term, in front of this term going to be? Essentially if you put it in this way it's going to be the third term that we actually care about. So what is this coefficient going to be? Now we have to clear, this coefficient, whatever we put here that we can use the binomial theorem to figure out isn't going to be this, this thing that we have to, I guess our actual solution to the problem that we posed is going to be the product of this coefficient and whatever other coefficients we have over here. Cause we're going to have 3 to the third power, six squared. So we're going to have to figure out what that is. But let's first just figure out what this term looks like, this term in the expansion. What this yellow part actually is. And there's a couple of ways that we can do that. We could use Pascal's triangle or we could use combinatorics. If we use combinatorics we know that the coefficient over here, this is going to be 5 choose 0, this is going to be the coefficient, the coefficient over here is going to be 5 choose 1. We've seen this multiple times. You could view it as essentially the exponent choose the the top, the 5 is the exponent that we're raising the whole binomial to and we say choose this number, that's the exponent on the second term I guess you could say. So this would be 5 choose 1. And this one over here, the coefficient, this thing in yellow. This is going to be 5, 5 choose 2. 5 choose 2. Now what is 5 choose 2? Well that's equal to 5 factorial over 2 factorial, over 2 factorial, times, times 5 minus 2 factorial. So let me just put that in here. Times 5 minus 2 factorial. And this is going to be equal to. Let's see 5 factorial is 5 times 4 times 3 times 2, we could write times 1 but that won't change the value. Over 2 factorial. Actually let me just write that just so we make it clear it is times 1 there. 2 factorial is 2 times 1 and then what we have right over here, this is 3 factorial, times 3 times 2 times 1. So let's see this 3 can cancel with that 3, that 2 can cancel with that 2, the 1's don't matter, won't change the value and then 4 divided by 2 is 2. So that is just 2, so we're left with 5 times 2 is equal to 10. So that's going to be this number right over here. This is going to be a 10. Now another we could have done it is using Pascal's triangle. We could have said okay this is the binomial, now this is when I raise it to the second power as 1 2 1 are the coefficients. When I raise it to the third power, the coefficients are 1, 3, 3, 1. When I raise it to the fourth power the coefficients are 1, 4, 6, 4, 1 and when I raise it to the fifth power which is the one we care about, the coeffiencients are going to be 1, 5, 10, 5 or sorry 10, 10, 5, and 1. And we know that when we go, this is going to be the third term so this is going to be the coefficient right over here. 1, 2, 3, third term. So either way we know that this is 10. But now let's try to answer our original question. What is this going to be? What are we multiplying times X to the sixth, Y to the sixth? And now we just have to essentially rewrite this expression. So it's going to be 10 times 3 to the third power, 3 to the third power, times Y squared to the third power, which is Y squared to the third power is Y to the sixth power. Y to the sixth power. Times six squared so times six squared times X to the third squared which that's X to the 3 times 2 or X to the sixth and so this is going to be equal to. Let's see it's going to be 10 times 27 times 36 times 36 and then we have, of course, our X to the sixth and Y to the sixth. So this is going to be, essentially, let's see 270 times 36 so let's see, let's get a calculator out. 270, I could have done it by hand but I'll just do this for the sake of time, times 36 is 9,720. So that's the coefficient right over here. It's going to be 9,720 X to the sixth, Y to the sixth. 9,720 X to the sixth, Y to the sixth and we're done.