Expansions (1 Viewer)

[KFunk]

Is there a general method for trinomial expansions (a + b + c )ⁿ [without using a 3d form of pascals triangle] or do I just need to remember the expansions for (a+b+c)³ and (a+b+c)⁴?

 

[Trev]

I never knew there was a 3D form of pascals triangle.
Anyone care to explain how it works?

 

[FinalFantasy]

for things like (a+b+c)³ i usually just make it ((a+b)+c)³
that (a+b+c)^n would be interesting, i would like to know a general way to do dat too!

 

[rama_v]

I think a similar question was asked at drmath.org: http://mathforum.org/library/drmath/view/51601.html

here's what the maths professor says: "We're already very close to coming up with a general formula for the coefficient of any term in the trinomial expansion. We just don't have a single symbol for the answer, equivalent to the "combinations" symbol."

 

[Trev]

Well that's no use!

 

[Xayma]

Is there a general method for trinomial expansions (a + b + c )ⁿ [without using a 3d form of pascals triangle] or do I just need to remember the expansions for (a+b+c)³ and (a+b+c)⁴?

A general way I don't think there is if you just want to find an individual term go with the grouping of two terms.

Consider:
(a+b+c)ⁿ

Let u=a+b

∴
(a+b+c)ⁿ
=(u+c)ⁿ

Suppose we want the term where the powers of a, b and c are h, i and j respectively.

Then we get:

ⁿC_ju^n-jc^j

But u=(a+b)

∴

ⁿC_j*(a+b)^n-j*c^j
=ⁿC_j*^n-jC_i*a^n-j-i*bⁱ*c^j
Since h+i+j=n

ⁿC_j*^n-jC_i*a^n-j-i*bⁱ*c^j
=ⁿC_j^n-jC_ia^hbⁱc^j

Now I suppose you could do this for all values of h, i and j if you really wanted to get an expansion.