Expansion with Pascal Triangle

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Homework Help Overview

The discussion revolves around expanding the expression (1+x+x²)⁵ using a method related to Pascal's Triangle. The original poster expresses confusion about how to approach this expansion, particularly since they are familiar with using Pascal's Triangle for binomial expansions but not for trinomials.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants suggest substituting variables to simplify the expression, such as replacing (1+x) with y. There is discussion about the implications of this substitution on the expansion process. Some participants also mention the multinomial theorem and express uncertainty about its application. Others propose expanding the expression by hand for small values of n to identify patterns.

Discussion Status

The conversation is ongoing, with various participants offering hints and alternative methods. Some have provided specific suggestions for approaching the problem, while others are exploring different interpretations of how to apply Pascal's Triangle to a trinomial expansion. There is no explicit consensus yet, but several productive ideas have been shared.

Contextual Notes

Participants note the complexity of expanding a trinomial compared to a binomial and question the assumptions underlying their approaches. There is also mention of potential resources, such as textbooks and online articles, that may provide further clarification on the topic.

Icetray
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Hi guys,

At a real loss here. I have a question that requires me to expend (1+x+x2)5 by constructing a pattern similar to that of the "Pascal Triangle". I understand the Pascal Triangle and know how to use with with expansions of two terms only (i.e. just (1+x)5)

Again, I have really have no clue how to get started and would really appreciate any hints that you guys can provide. (I would really like to solve this myself if possible (: )

Thanks guys!
 
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Try substituting y at the the place of (1+x). :)
 
Pranav-Arora said:
Try substituting y at the the place of (1+x). :)

Thanks for you speedy reply! (: Anyway do you mean expend (y+x2)5 instead? Would that not mean that when I re-substitute y = 1+x into the answer, I'll have more expansions to carry out?

Thanks again!
 
Yes, and after you have done that you will have, of course, terms of y up to y^5 so you will need to use Pascal's triangle to exand each power of y= 1+ x.
 
HallsofIvy said:
Yes, and after you have done that you will have, of course, terms of y up to y^5 so you will need to use Pascal's triangle to exand each power of y= 1+ x.

Alright! I'll go and try it out. (: Thanks for the help Pranav-Arora and HallsofIvy! Much appriciated! (:
 
Icetray said:
Alright! I'll go and try it out. (: Thanks for the help Pranav-Arora and HallsofIvy! Much appriciated! (:

Don't go that way now, i think i have found a much simpler way for it. (But i am also not able to understand it). Check out the multinomial theorem on Wikipedia:-
http://en.wikipedia.org/wiki/Multinomial_theorem

I don't understand what does the sigma notation mean here?
 
Pranav-Arora said:
Don't go that way now, i think i have found a much simpler way for it. (But i am also not able to understand it). Check out the multinomial theorem on Wikipedia:-
http://en.wikipedia.org/wiki/Multinomial_theorem

I don't understand what does the sigma notation mean here?

I was thinking that there might be an easier way as well. All that expending seems like it'll take much effort. And I visited the link that you provided. That's way to complicated. Anyone knows if there's anything on the Pascal Triangle in Thomas' Calculus? I'm currently using that as my reference book.
 
Pranav-Arora said:
But i would recommend you to learn multinomial theorem. I too don't know it so I have started a thread on multinomial theorem, you can follow that to learn it. Link:-
https://www.physicsforums.com/showthread.php?p=3449686#post3449686

Thanks for everything Pranav! (: I look forward to replies on your thread! (: Anyway I realized that for this question, they want us to use a pattern similar to that of the Pascal Triangle. I assume that in this case, substituting y=1+x is the only way? ):
 
  • #10
Since there are three terms (a trinomial), use http://en.wikipedia.org/wiki/Pascal's_pyramid#Structure_of_the_Tetrahedron".

Added in Edit:

Another idea: (But some of my ideas are half-baked at best.)
Use Pascal's Triangle with the following.

1 + x + x2 = 1 + 2x + x2 - x = (1 + x)2 - x

Therefore, (1 + x + x2)5 = [(1 + x)2 - x]5
= [(1 + x)2]5 + 5[(1 + x)2]4x + 10[(1 + x)2]3x2 + 10[(1 + x)2]2x3 + 5[(1 + x)2]x4 + x5


Yup, still pretty messy !
 
Last edited by a moderator:
  • #11
Here are two hints.

First, expand (1 + x + x^2)^n "by hand" for a few small values of n-- say at least n = 1,2,3.

Second, since you are dealing with a trinomial instead of a binomial, expect to add 3 numbers instead of 2 to get each new entry in the triangle.

Here is a start...

Code:
            1
        1   1   1
    1   2   3   2   1
      ...   7   ...
 
  • #12
awkward said:
Here are two hints.

First, expand (1 + x + x^2)^n "by hand" for a few small values of n-- say at least n = 1,2,3.

Second, since you are dealing with a trinomial instead of a binomial, expect to add 3 numbers instead of 2 to get each new entry in the triangle.

Here is a start...

Code:
            1
        1   1   1
    1   2   3   2   1
      ...   7   ...

Hey guys. Just came back from my tutorial. What my professor actually did was what the other members suggested her except that he let 1= a and x+x2=b. From there he used the pascal triangle to expend out (a+b)5. Where he later used the pascal triangle to expend out the remaining factors again. (:

Anyway thank you so much for all your help guys! :D
 
  • #13
awkward said:
Here are two hints.

First, expand (1 + x + x^2)^n "by hand" for a few small values of n-- say at least n = 1,2,3.

Second, since you are dealing with a trinomial instead of a binomial, expect to add 3 numbers instead of 2 to get each new entry in the triangle.

Here is a start...

Code:
            1
        1   1   1
    1   2   3   2   1
      ...   7   ...
awkward has a great idea here.

(Add the number directly above to the number on either side of the number above. Any (missing) number to the left or right of the triangle is taken to be zero.)

Let's complete the 4th row. (It's for (1+x+x2)3.)

Code:
            1
        1   1   1
    1   2   3   2   1
1   3   6   7   6   3   1

So (1+x+x2)3 = 1 + 3x + 6x2 + 7x3 + 6x4 + 3x5 + x6 .
 
  • #14
I think this is answer

1
5 5
10 20 10
10 30 30 10
5 20 30 20 5
1 5 10 10 5 1
 

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