Binomial formula

  1. tony873004

    tony873004 1,585
    Science Advisor
    Gold Member

    1. The problem statement, all variables and given/known data
    Find the term with the specified power in the expansion of the given binomial power.
    [tex]
    \left( {x^3 + y^2 } \right)^{42} ,\,\,\,\,\,y^{15}
    [/tex]


    2. Relevant equations
    [tex]{\rm{term}} = \frac{{n!}}{{r!\left( {n - r} \right)!}}x^{n - r} y^r [/tex]



    3. The attempt at a solution
    [tex]\begin{array}{l}
    {\rm{term}} = \frac{{42!}}{{15!\left( {42 - 15} \right)!}}x^{3 \cdot \left( {42 - 15} \right)} y^{2 \cdot 15} \\
    \\
    {\rm{term}} = \frac{{42!}}{{15!\left( {27} \right)!}}x^{81} y^{30} \\
    {\rm{term}} = {\rm{98672427616}}\,x^{81} y^{30} \\
    \end{array}
    [/tex]

    The back of the book says no such term exists. Why? Is it because x has an exponent that is higher than n? x^3 doesn't have a higher exponent, and I thought that's all that mattered.

    Also, is there a way of simplifying that factorial so I don't have to rely completely on the calculator to solve? Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. I would think a [itex]y^{15}[/itex] term does not exist because every [itex]y[/itex] term is raised to an even power? I've never used the formula before to calculate the term with the given binomial power so I can't comment on that part if you did it correctly.

    For this factorial there isn't much else you can do except rewrite [itex]\displaystyle\frac{42!}{15! \cdot 27!}[/itex] as [itex]\displaystyle\frac{(42 \cdot 41 \cdot \cdot \cdot 28)27!}{15! \cdot 27!} = \frac{(42 \cdot 41 \cdot \cdot \cdot 28)}{15!}[/itex] and then cancel out like terms to get rid of the 15!.
     
  4. tony873004

    tony873004 1,585
    Science Advisor
    Gold Member

    Thanks! The longer I stared at this, I started to realize that there's nothing I can multiply by 2 that will give me 15. So I imagine that "does not exist" is also the answer to[tex]\left( {x^3 + y^2 } \right)^{42} ,\,\,\,\,\,y^{15}[/tex]
    This is an even question, so no back of book answer.
     
  5. Well, no number in the inters to multiply 2 by to get 15! :P

    I think you recopied the problem from the first post, but did it involve another instance where you can't multiply an integer to get the specified power?
     
  6. tony873004

    tony873004 1,585
    Science Advisor
    Gold Member

    oops, my copy and paste skills need improving!
    [tex]\left( {x^3 + y^2 } \right)^{107} ,\,\,\,\,y^{77} [/tex]
     
  7. Yeah, same type of case.
     
  8. tony873004

    tony873004 1,585
    Science Advisor
    Gold Member

    Going back to the first example, if, rather than use the formula, I decide to actually waste a few sheets of paper expanding this this thing, my first few terms will be
    [tex]
    x^{3\left( {42} \right)} y^{2\left( 0 \right)} + x^{3\left( {41} \right)} y^{2\left( 1 \right)} + x^{3\left( {40} \right)} y^{2\left( 2 \right)} + x^{3\left( {39} \right)} y^{2\left( 3 \right)} + ...
    [/tex]
    which is pretty much all I need to tell me that y^15 will never happen.

    Thanks for the late night help!
     
  9. Don't forget to use pascal's triangle to put the correct coefficients in front of the terms! But, yep, that's why there is no term!
     
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