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Expansion with Pascal Triangle

  1. Aug 13, 2011 #1
    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!
     
  2. jcsd
  3. Aug 13, 2011 #2
    Try substituting y at the the place of (1+x). :)
     
  4. Aug 13, 2011 #3
    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!
     
  5. Aug 13, 2011 #4

    HallsofIvy

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    Yes, and after you have done that you will have, of course, terms of y up to [itex]y^5[/itex] so you will need to use Pascal's triangle to exand each power of y= 1+ x.
     
  6. Aug 13, 2011 #5
    Alright! I'll go and try it out. (: Thanks for the help Pranav-Arora and HallsofIvy! Much appriciated! (:
     
  7. Aug 13, 2011 #6
    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?
     
  8. Aug 13, 2011 #7
    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.
     
  9. Aug 13, 2011 #8
  10. Aug 13, 2011 #9
    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? ):
     
  11. Aug 13, 2011 #10

    SammyS

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    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: Apr 26, 2017
  12. Aug 14, 2011 #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 (Text):

                1
            1   1   1
        1   2   3   2   1
          ...   7   ...
     
     
  13. Aug 16, 2011 #12
    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
     
  14. Aug 16, 2011 #13

    SammyS

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    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 (Text):

                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 .
     
  15. Dec 4, 2011 #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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