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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Toothpaste on October 31, 2007, 10:03:32 pm

Title: Pascal's triangle
Post by: Toothpaste on October 31, 2007, 10:03:32 pm

I've never learned about Pascal's triangle.

Do I need it?

Title: Pascal's triangle
Post by: Ahmad on October 31, 2007, 10:13:57 pm

Not for methods, by the looks of it. Although it's very important in combinatorics, so I would learn it! Try to find some cool identities using it  :wink:

Title: Pascal's triangle
Post by: Khangfu on October 31, 2007, 10:17:38 pm

Who needs pascals when you got the binomial theorem :P

Title: Pascal's triangle
Post by: enwiabe on October 31, 2007, 10:26:56 pm

o.O yes you do? I remember pascal's triangle questions last year for when you expand out using the binomial theorem.

Title: Pascal's triangle
Post by: Ahmad on October 31, 2007, 10:30:17 pm

Unless you use the algebraic definition of binom..

Title: Pascal's triangle
Post by: Toothpaste on October 31, 2007, 10:35:24 pm

Guess what, I've never learned the binomial theorem.

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Title: Pascal's triangle
Post by: enwiabe on October 31, 2007, 10:37:17 pm

And that's just one of the reason's why you'll get <40 for Methods. :D

Title: Pascal's triangle
Post by: Toothpaste on October 31, 2007, 10:38:39 pm

Yeah, probably. Fuck.

Title: Pascal's triangle
Post by: asa.hoshi on October 31, 2007, 10:40:31 pm

dont worry.

if its on exam 1 then u are stuffed...
but if its on exam 2... u have ur calc to calc. it.

Title: Pascal's triangle
Post by: Ahmad on October 31, 2007, 10:41:59 pm

Learn it now, it's nothing huge. It'll take less than half an hour, promise.  :)

Title: Pascal's triangle
Post by: Toothpaste on October 31, 2007, 10:44:16 pm

... and I was all set to do absolutely nothing till next Friday.

Title: Pascal's triangle
Post by: Collin Li on October 31, 2007, 11:04:43 pm

It's not on the course. You don't need to know how to find the coefficient of x^4 for (3x+2)^7.

Title: Pascal's triangle
Post by: Toothpaste on October 31, 2007, 11:07:24 pm

Rejoice!

(But I'll learn it since I don't want to be left out of maths world.)

Title: Pascal's triangle
Post by: Collin Li on October 31, 2007, 11:11:42 pm

Quote from: "Toothpick"

Rejoice!

(But I'll learn it since I don't want to be left out of maths world.)

It's pretty simple:

Code: [Select]

  1 1
  1 2 1
 1 3 3 1
1 4 6 4 1
   ...

Notice how the numbers from the row above add up together to give the number below? That is Pascal's triangle.

The binomial theorem links to this by saying that you can generate any number from this triangle using nCr, where:
n = the row, and
r+1 = the column
...on Pascal's triangle.

This links into polynomial expansion:

For example:
(x+a)^3 = x^3 + 3ax^2 + 3a^2x + a^3

Notice the coefficients? (1, 3, 3, 1) This is row 3 of Pascal's triangle, for expanding a polynomial of degree 3.

Now, I think you're clever enough to figure out how to find out the coefficient to x^4 for (2x+3)^7

Title: Pascal's triangle
Post by: Toothpaste on November 01, 2007, 12:01:26 am

Hah, thanks!

Hmm..
(2x+3)^7 ..

so the coefficient of x^4 is

(7 C 4) (2x)^4 3^(3)

15120?

Title: Pascal's triangle
Post by: Collin Li on November 01, 2007, 12:06:14 am

Quote from: "Toothpick"

(2x+3)^7 ..

so the coefficient of x^4 is

(7 C 4) (2x)^4 3^(3)

15120?

Ehh, yeah, but only by luck:

x^7, x^6, x^5, x^4

x^4 is the 4th term, which means we're looking at column 4, so r+1=4 => r=3.

7C3 is what you needed, but 7C4 = 7C3 because of the identity: nCk = nC(n-k)

Try again if I had asked for the x^5 term?

Title: Pascal's triangle
Post by: Toothpaste on November 01, 2007, 12:15:51 am

7-r = 5   (is this valid? It works lol)
r=2

but yeah it's the 3rd term so r +1 = 3 , r=2

coefficient of x^5 would be

(7 C 2) (2x)^5 3^(2)

6048

Title: Pascal's triangle
Post by: Toothpaste on November 01, 2007, 12:31:43 am

Yeah, it's clear now.


(2x+3)^7
= (2x)^7 + 7(2x)^6 3 + 21 (2x)^5 3^2 ... etc

:D

Title: Pascal's triangle
Post by: Collin Li on November 01, 2007, 12:46:59 am

Quote from: "Toothpick"

7-r = 5   (is this valid? It works lol)
r=2

but yeah it's the 3rd term so r +1 = 3 , r=2

coefficient of x^5 would be

(7 C 2) (2x)^5 3^(2)

6048

Yeah, using n-r = k'th term works fine as well. I think that's the proper definition anyway, my r+1 = k'th term happens to work as well, though :P