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Binomial formula

by tony873004
Tags: binomial, formula
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tony873004
#1
Mar13-12, 12:55 AM
Sci Advisor
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P: 1,542
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
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scurty
#2
Mar13-12, 01:28 AM
P: 392
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!.
tony873004
#3
Mar13-12, 01:37 AM
Sci Advisor
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P: 1,542
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.

scurty
#4
Mar13-12, 01:42 AM
P: 392
Binomial formula

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?
tony873004
#5
Mar13-12, 02:22 AM
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P: 1,542
oops, my copy and paste skills need improving!
[tex]\left( {x^3 + y^2 } \right)^{107} ,\,\,\,\,y^{77} [/tex]
scurty
#6
Mar13-12, 02:30 AM
P: 392
Yeah, same type of case.
tony873004
#7
Mar13-12, 02:38 AM
Sci Advisor
PF Gold
P: 1,542
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!
scurty
#8
Mar13-12, 02:49 AM
P: 392
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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