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## Precalculus

### 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? •  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.
• What if there were coefficients on the variables? •  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
• Hey Sal, does this method only work for addition binomial? I can't do the same thing with subtraction in it can I? • I want to ask that is this algolthm rule applicable for even exponents ?If yes then how to apply symmtery? • How do we expand the powers like 1/3 or 1/2? • How to find general and middle term in binomial expansion? • 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! • 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?   