[Discrete Mathematics] Binomial Theorem and Pascal's Triangle

In this video, we look at the Binomial Theorem and how it relates to Pascal's Triangle.

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[BiranchiNarayanNayak]

Awesome i just loved the Binomial theorem using pascal triangle....
I had never learnt in my high school course...

[dianampm]

This is a really good video! I wasnt hoping to learn all the things I needed in one video tbh, and ended up closing all the other tabs i thoguth id had to watch. thanks!!

[rafyyd]

thank you so much! binomial theorem using pascal's triangle blew my mind

[hannahweiss]

hope this saves my grade thank u so much

[LaneSurface]

Grimaldi gives an interesting proof of the binomial theorem. He says that in the expansion of (x+y)^n=(x+y)(x+y)...(x+y), then for the term x^k*y^(n-k), there are C(n, k) ways to select k x's and C(n, n-k) ways to select n-k y's. Since C(n, k)=C(n, n-k), that is the number of repeated x^k*y^(n-k) terms (or, in other words, its coefficient). Thus we can express the k-th term as C(n, k)*x^k*y^(n-k). You can also do this by induction, but it's not as simple as the combinatorial proof.

[skjbanks]

When you first go over Pascal's triangle you begin the numbering with a zero, then you start with a one with your first example

[zakariabenjelloun]

nice video i love it thank you you explain very well

[rkcst]

@7:24
Where did the (-b)^1 go???
You just treated it as if it was equal to -1... Is that correct? I must be missing something. Is any negative number raised to 1 going to be -1?

[zYaaaaan]

hey....
How do you get 1(16aPOWER4)?

[cqlpvte]

Interesting thing about the values from Pascals Triangle and the powers is, at least in the example given at with (x+y), each leading value is like taking the derivative of x and integral of y at the same time. When we take the derivative we multiple the lead value and subtract 1 from the power, and when we take an integral we add 1 to the power and divide the leading value by that new power number.

Example:

1x^6 (derivative of x^6 is 6x^5, integral of y^0 is 1/1*Y^1) so next in line is 6/1 x^5y^1,

then do it again for 6x^5y^1, derivative of x^5 is 5x^4, integral of y^1 is 1/2*y^2, so 6 * 5 *1/2 is 15 -> 15x^4y^2

15x^4y^2, derivative of x^4 -> 4x^3, integral of y^2 -> 1/3*y^3, 15 * 4 * 1/3 = 20 -> 20x^3y^3

20x^3Y^3 -> derivative of x^3 -> 3x^2, integral of y^3 -> 1/4*y^4, 20 * 3 * 1/4 = 15 -> 15x^2y^4

15x^2y^4 -> derivative of x^2 -> 2x, integral of y^4 -> 1/5 * y^5 -> 15 *2*1/5 =6 -> 6xy^5

6xy^5 -> derivative of x -> 1, integral of y^5 -> 1/6 * y^6, 6*1*1/6 = 1 -> 1y^6

Add them up and you get x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6

Don't think there is some correlation or helpful usage of this, just something that was interesting that I noticed.

[jonathanmalmberg]

There's something I don't quite understand at ca 3:40. Why do you get 0! when 2 chooses 2? Don't we use the formula
n!/k!(n-k)! ? And why isn't 4 chooses 4 0! ?

[aboogiewithdahoodie]

13:03 Starting with 0, still on coding/programming 😂

[drainedgaia]

4:50
I perceive it as plugging in each term of the equation as whole, when dealing with the binomial theorem.
14:21
You almost had me right there :)

[sunnyzhu]

i can understand 70%, who understand 100%

[joecremeno]

your theorem is reversed, you did not make me happy.