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### Course: Precalculus > Unit 9

Lesson 3: The binomial theorem# Expanding binomials w/o Pascal's triangle

Sal gives a "trick" for expanding large powers of binomials, without using Pascal's triangle. Created by Sal Khan.

## Want to join the conversation?

- How would you prove this algorithm? Would you do a regular concrete proof or a proof by induction?(44 votes)
- In accordance with the Binomial Theorem a coefficient equals to
*n!*/(k!(n-k))!

Sal has shown us that it is also possible to find a coefficient in another way.

It is known that n is a constant throughout the whole expression and k changes at every term (*k=0*at the first term,*k=1*at the second term, etc.). Let's say that k of the term for which we want to calculate the coefficient is K. Then k of the previous term is*(K-1)*.

Sal says that the coefficient of the current term equals to the exponent of the previous term (i.e.*n - k = n - (K-1) = n - K + 1*) terms the coefficient of the previous term (i.e.*n!*/(k!(n-k)!) =*n!*/((K-1)!(n - (K-1))!) =*n!*/((K-1)!(n - K + 1)!) ), divided by the sequence number of the previous term (actually, this equals to*k + 1 = (K-1) + 1 = K*). So we get the following expression:*((n!(n-K+1)) / ((K-1)!(n- K+1)!))*/ K.

Let's simplify it:*((n!(n-K+1)) / ((K-1)!(n- K+1)!))*/ K =*(n!(n-K+1))*/ (K(K-1)!(n-K+1)!) =*n!*/ (K(K-1)!(n-K)!) =*n!*/ (K!(n-K)!).

So we've got the standard form of the coefficient in accordance with the Binomial Theorem.(27 votes)

- What if there were coefficients on the variables?(16 votes)
- Great question.

You would treat the coefficient + the letter variable of the term as a new variable itself.

For example: (2x + 3y)^2

(2x)^2 + 2(2x)(3y) + (3y)^2

Then you do your order of operations or combining like terms

4x^2 + 12xy + 9y^2

You can also use substitution to help ease the confusion.

For example: (2x + 3y)^2

You can say lets 2x=a and 3y=b

Then rewrite it to (a + b)^2

a^2 + 2ab + b^2

Now you substitute back in the values for a and b

(2x)^2 + 2(2x)(3y) + (3y)^2

4x^2 + 12xy + 9y^2(40 votes)

- Hey Sal, does this method only work for addition binomial? I can't do the same thing with subtraction in it can I?(4 votes)
- Yes you can, every 2nd term becomes negative, for example: http://www.wolframalpha.com/input/?i=%28a-b%29%5E5 (check expanded form)(16 votes)

- I want to ask that is this algolthm rule applicable for even exponents ?If yes then how to apply symmtery?(4 votes)
- Yes, it still applies. The even exponents (0, 2, 4, 6...) are the first, third, fifth, seventh, etc rows of the triangle. The even exponent rows have a single highest value in the middle. The odd exponent rows how a pair of highest values. See a picture: http://en.wikipedia.org/wiki/Pascal's_triangle(6 votes)

- How do we expand the powers like 1/3 or 1/2?(7 votes)
- Actually, when you raise it to a fraction you're just calculating the root of the expression.

So, (a+b)^(1/2) is the same as sqrt(a+b) and (a+b)^(1/3) is the same as the cubic root of (a+b).

The numerator is the actual power which you're raising the expression to, the denominator is just the root's index.(2 votes)

- How to find general and middle term in binomial expansion?(4 votes)
- I was struggling to understand the relationship between Pascal Triangle, combinations and the binomial theorem all day until I remembered the Probability function and coin tosses where we toss coins and count all of the possible events of getting 2 heads and 1 tail in 3 tosses for example. Once I remembered that, it was quite easy to understand using that analogy. Pascal triangle is the same thing. Binomial Theorem is composed of 2 function, one function gives you the coefficient of the member (the number of ways to get that member) and the other gives you the member. The coefficient function was a really tough one. Pascal and combinations. Seems logical and intuitive but all to nicely made. In triangle theorem you will start to see a pattern of repeating numbers. Such as row 3 of pascal triangle where you have 2 coefficients that are same and it goes on in the next rows. This happens because, for example, combination of 3 elements with 2A´s (one must treat A´s as 2 seperate elements) to choose from (your member elements being a²b = AAB) is the same as choosing 1 B out of AAB into 3 slots. This is why this happens. AAA and BBB happen because, if thinking in terms of A: Number of way to choose 3 A´s from element member AAA and put into 3 slots (3C3) is equal to thinking in terms of B where you ask yourself to get to AAA, how many B´s must I choose? Answer is 3C0. 3C0 = 3C3. You can think either in terms of A´s or B´s. And there you have it. Simple coin tossing analogy, as simple as that!(4 votes)
- In the video's example, Sal uses the previous coefficient and exponent to find out the terms for the problem. Will this always be the case or did he just uses this for his example?(3 votes)
- Based on the other answers here, I think this does work for other problems as well.(2 votes)

- what is the formula for (x+y) to the nth power?(2 votes)
- Sum of all the terms with the format

nCr * x^(n-r) * y^r.

nCr is n Choose r, which is equal to n!/[(n-r)!(r)!]

nCr * x^(n-r) * y^r is also the formula for the r+1 term of the expansion.(2 votes)

- How do I find a specific term for binomial expansion? This was covered in my additional maths class but not on Khan Academy. Can anyone please give me a hand? Thanks!(2 votes)

## Video transcript

Voiceover:What I want to
show you in this video is what could be described
as, I guess, a trick for finding binomial expansions, especially binomial expansions where
the exponent is fairly large. But what I want you to do after this video is think about how this
connects to the binomial theorem and how it connects to Pascal's Triangle. Now let me show you the trick. I'm going to take (X+Y)^7. That's going to have eight terms. How do I know that? Well (X+Y)^1 has two
terms, it's a binomial. (X+Y)^2 has three terms. (X+Y)^3 has four terms. So this is going to have eight terms. Let me just create little
buckets for each of the terms. This is the bucket, these aren't the coefficients these are just the buckets. 1st term, 2nd term, 3rd term, 4th term, 5th term, 6th term,
7th term, and 8th term. Now lets write out the actual X's and Y's. The first term, we're
going to start with X^7. Then each term after that our degree or our power on the X goes down by one. So you go X^6, X^5, X^4, X^3, X^2, X^1, we could just write that as X, and this is going to be X^0, which
is just going to be one. Now lets think about Y. This is going to start at Y^0, which is just one so I'm
not going to write it, Then it's going to be Y^1,
Y^2, Y^3, Y^4, Y^5, Y^6, and then Y^7 and you can
verify you got it right because for each term, the
exponents should add up to seven. You see that even here. This is X^1 times Y^6. Those add up to be seven. Now, lets get to the interesting part, which is actually
calculating the coefficient. And the algorithm is for
each term right over here, so lets just start, we
know the coefficient right over here is going to be one. Actually, let me write that down. The coefficient over
here is going to be one. So for each term, the
coefficient right over here, I'm going to try to color-code
it so we can see it, the coefficient is
going to be the exponent of the previous term, so the
exponent of the previous term in this case is the seven, the
exponent of the previous term, times the coefficient of the
previous term, divided by which term that actually
was, so divided by the term. That was the 1st term,
so now the coefficient on the 2nd term is seven
times one divided by seven, which is going to be equal to seven. Now what about this one? We use the exact same process. The exact same process. It is going to be the exponent
on the X term, I guess you could say the exponent on
X, I guess we could go with. The exponent on the X, which is six, times the coefficient of the previous term, so times seven, so we're
taking the X power times the coefficient of the
previous term, so times seven. So the X part of the previous term, times the coefficient
of the previous term, divided by the actual,
I guess you could say, index of the previous
term, so divided by two. So what is that going to be? So this is the equivalent
to three times seven, so this is going to be equal to 21. And now lets go to this
term, exact same idea. Go to the previous term,
what's our exponent on X? It is five. Lets multiply it times the coefficient, so lets multiply it times
21, and then lets divide it by which term that is,
so that was the 3rd term. So this is going to be, lets see, five times 21 over three is seven, so this is going to be
35, five times seven. And we can keep going, or we can recognize that there's a symmetry here. If this is one, then the last
term is also going to be one. If the second term is seven, then the second-to-last term is seven. If the third term is 21, then the third term to the last is 21. And then if the 4th term is 35, then the fourth from the last is 35. And just like that, we have figured out the expansion of (X+Y)^7. Pretty neat, in my mind.