# Algorithm for mentally computing binomial expansion coefficients | Algebra II | Khan Academy

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.

## 21 thoughts to “Algorithm for mentally computing binomial expansion coefficients | Algebra II | Khan Academy”

1. elbay2 says:

Neat trick, and well explained!

2. HobbesGraal says:

Doesnt seem to work with problems like (x^2-4)^4

3. matthew duke says:

did you develop this?

4. Sir Trollface says:

This is a lot more efficient than using the binomial theorem formula. I really needed this type of method since I prefer calculating mentally rather than using paper.

5. Sahil Bodhwani says:

i love dis

6. Anikash Chakraborty says:

this is simply C(x,y)

7. Randrita Sarkar says:

awsm 1……..really awsm trick

8. Hacker Man 108 says:

Wow!!
It's way better than most of the methods out there!

9. tota aliraqia says:

thank you mr. khan πππ

10. verniel love says:

AMAZINGπππππππππ€… Once again math has gained my loveπΉπΉπΉβ€οΈ thanks a bunch Mr.Khan

11. Simon Zumbach says:

How did he figure this out??

12. Appleberry Smith says:

That last pun… I see what you did there.

13. metal9lover9maniac says:

Thanks. This is helpful so far, but I don't know if there's a problem with my flash player, because on my end the audio kept repeating a few times. This method the most useful I've seen so far.

14. Nathan Nelson says:

Mind blown!!!

15. medievalmusiclover says:

Pretty Neat! Like this word! Excellente explanation and method. You are wonderful natural Professor!

16. Manish Vyas says:

Why this algorithm works

17. Yassin Ibrahim says:

Thank You very much for all your videos.AMAZING!

18. wagun evans says:

this isn't a magic, but just a long method instead

19. GENERAL CUBER says:

What if there are odd no. Of terms in expression

20. annoyingkraken says:

First time I saw this technique. Love it.

21. Brigid Flynn says:

Yo thats whack . Im shook