(a + b + c.....)^nth???

E

edt966

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Dec 15, 2005
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I've been playing around with (a + b + c)^n and have gotten up to the 5th power looking for a pattern, there seems to be one but my eyes are too buggy to go on. Is there a formula for uglies like these? I found a good short cut for binomials to the nth power, but would sure die for a short cut (not really) for (a + b + c +...)^nth power. I've got a learning resource at school to help with verification and nobody seems to know of a formula. I'm seeing a pattern of some sort emerge in the leading coefficiants, but can't nail anything down

Eddie :lol:
 
Yep, tried Pascals triangle. no go, and the binomial theorem (at least the short cut I made up) works only on quotents of 2 terms (a + b). I'll check out that other one you mentioned. Thanks.
 
First one needs to understand that the sum must be finite!
That is something such as \(\displaystyle \L
\left( {\sum\limits_{k = 1}^n {x_k } } \right)^N = \left( {x_1 + x_2 + \cdots + x_n } \right)^N\).

Each term in the expansion looks like:

\(\displaystyle \L
\frac{{N!}}{{\left( {\alpha _1 } \right)!\left( {\alpha _2 } \right)! \cdots \left( {\alpha _n } \right)!}}\left( {x_1 } \right)^{\alpha _1 } \left( {x_2 } \right)^{\alpha _2 } \cdots \left( {x_n } \right)^{\alpha _n }\)

here each \(\displaystyle \L
{\alpha _k }\) is an integer, \(\displaystyle 0 \le \alpha _k \le n\) and \(\displaystyle \alpha _1 + \alpha _2 + \cdots \alpha _n = N\).

There are \(\displaystyle \L
\frac{{\left( {N + n - 1} \right)!}}{{\left( {N!} \right)(n - 1)!}}\) terms in the expansion.


Example, expand \(\displaystyle \L
\left( {a + b + c} \right)^{5}\) and there would be 21 terms.

Here are three of the terms:
\(\displaystyle \L
b^5\), \(\displaystyle \L
\frac{{5!}}{{\left( {2!} \right)(2!)(1!)}}a^2 b^2 c\), and \(\displaystyle \L
\frac{{5!}}{{\left( {2!} \right)(3!)}}a^2 c^3\)
 
Thanks, I think I got it with your help!

(a + b + c) ^5

a^5*5!/5! next a^4b 5!/4!1!=5/1! etc.....a^3b^2......a^2b^2c...
b^5 a^4c
c^5 b^4a
b^4c
c^4a
c^4b


and if there is coefficiants then a's coefficiant would go to the 5th power and times 5!/5! which is 1 and so on for b^5 and c^5. Then a^4b would be a's coefficiant to the 4th * b's coefficiant to the once power * 5 for that column. Makes sense to me...really.

eddie.
 
Hello, Eddie!

I too was looking for a pattern (years and years ago) and finally found one.

The problem was that the terms were not nicely ordered.

\(\displaystyle n\,=\,1:\;\;(a\,+\,b\,+\,c)^1\:=\:a\,+\,b\,+\,c\)

\(\displaystyle n\,=\,2:\;\;(a\,+\,b\,+\,c)^2\:=\:a^2\,+\,2ab\,+\,2ac\,+\,b^2\,+\,2bc\,+\,c^2\)

\(\displaystyle n\,=\,3:\;\;(a\,+\,b\,+\,c)^3\:=\:a^3\,+\,3a^2b\,+\,3ab^2\,+\,b^3\,+\,3b^2c\,+\,3bc^2\,+\,c^3\,+\,3a^2c\,+\,3ac^2\,+\,c^3\)


I labored to get them the terms into some logical order
\(\displaystyle \;\;\)and saw that a triangular array was involved.
[Ignore the dots ... they were used for spacing.]

For \(\displaystyle n\,=\,1:\;\;\begin{array}{cc} . & a & .\\ b & . & c\end{array}\)

For \(\displaystyle n\,=\,2:\;\;\begin{array}{ccc}. & . & a^2 & . & .\\ . & 2ab & . & 2ac & . \\ b^2 & . & 2bc & . & c^2\end{array}\)

For \(\displaystyle n\,=\,3:\;\;\begin{array}{cccc}. & . & . & a^3 & . & . & . \\ . & . & 3a^2b & . & 3a^2c & . & . \\ . & 3ab^2 & . & 6abc & . & 3ac^2 & .\\ b^3 & . & 3b^2c & . & 3bc^2 & . & c^3 \end{array}\)


If you "stack" these triangle, you get "Pascal's Tetrahedron".
\(\displaystyle \;\;\)Each coefficient is the sum of the three coefficients directly above it.
 
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