Question

How to do the expansion of \(\large(a + b + c)^n\)?

Answer 1

You need Multinomial theorem.

http://en.wikipedia.org/wiki/Multinomial_theorem

Answer 2

Wait, I'll just return and try....you may stop viewing the question :D

Answer 3

use multinomial theorem

Answer 4

Hmmm...this is perhaps too difficult for me....can you guys make it easier, or explain?

Answer 5

here's a example @ParthKohli http://in.answers.yahoo.com/question/index?qid=20100125032321AA2FDqI

Answer 6

@Glen_McGrath please stop posting irrelevant things here

Answer 7

@ParthKohli it's not irrelevant

Answer 8

I want to learn it.

Answer 9

Can you teach it to me?

Answer 10

its just an example

Answer 11

well in startng frm whr u copied it means u r unable to answer

Answer 12

Coefficients in a Trinomial Expansion basically its a concept u r trying to solve try reading the book as I explain it would really be diff...

Answer 13

I learnt about the coefficients in binomial expansion...but what about trinomial?

Answer 14

Yes, this is very closely related to the binomial
theorem, Pascal's triangle, and "m choose n."

Let's look first at the more familiar binomial theorem, Pascal's
triangle, and the expansion of\[ (a+b)^n. \]

As you know, the expansion of \[(a+b)^n\] is:

\[ (a+b)^n = C(n,n)*a^n + C(n,n-1)*a^(n-1)b + ... + C(n,0)*b^n \]

The coefficient of the a^(n-k).b^k term in this expansion is

n!
C(n,n-k) = ---------
(n-k)!*k!

Answer 15

posting very hard formulae is fun ^_^

Answer 16

Lol I don't need to know about the binomial one....

Answer 17

here comes Fool For Math And Lg

Answer 18

they are GREAt HELPERS

Answer 19

\[1 + n Log(a + b + c) + [(1/2) n^2] Log^2(a + b + c) + [(1/6) n^3]Log^3(a + b + c)\]

\[[(1/24)n^4]Log^4(a + b + c) + [(1/120)n^5]Log^5(a + b + c) + O(n^5)\]

Answer 20

Okay this is probably easier: \(\large(a + b + c)^n = \large( (a + b) + c)^n\)

Now apply binomial theorem.

Answer 21

means I'm right

Answer 22

answer = 1ac+0a-1b+

Answer 23

Some thing happened to my keyboard :/

Answer 24

I'm trying to press backslash and it's coming out as #.

Answer 25

Question answered. I'm gong to finish my work now. Adios.

Answer 26

Uh, here's an attempt of explaining:

\[(x_1 + x_2 + \cdots + x_m)^n
= \sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m}
\prod_{1\le t\le m}x_{t}^{k_{t}}\]
This is a bit complex, but your case is the specific parameters:
\(x_1=a, \quad x_2=b, \quad x_3=c, \text{ and } m=3\)

So that big formula just means, in your case,
\[(a+b+c)^n=\sum_{k_1+k_2+k_3=n}{n \choose k_1, k_2, k_3}
\prod_{1\le t\le 3}x_{t}^{k_{t}}\]
Interpreting this notation is somewhat difficult. First of all,
\[\prod_{1 \leq t \leq 3}x_t^{k_t}=x_1^{k_1}x_2^{k_2}x_3^{k_3}=a^{k_1}b^{k_2}c^{k_3}\]
So we can say:
\[(a+b+c)^n=\sum_{k_1+k_2+k_3=n}{n \choose k_1, k_2, k_3}a^{k_1}b^{k_2}c^{k_3}\]
Now, the thing to realize here is this: The summation notation is abstract; it is not an obvious notation. Wiki states, "The sum is taken over all combinations of nonnegative integer indices \(k_1\) through \(k_m\) such that the sum of all \(k_i\) is \(n\). That is, for each term in the expansion, the exponents of the \(x_i\) must add up to \(n\). "

This is where things get really complicated for me...

The idea here is this: The sum is evaluated at all triplets \((k_1,k_2,k_3)\) such that \(k_1+k_2+k_3=n\). This is a large set of triplets. Talking more detailedly about this fails at this exact point because there isn't (that I can see) a more straightforward way of explaining a *general* statement.

But here is a brief example. Let \(n=2\). All \((k_1,k_2,k_3)\) such that \(k_1+k_2+k_3=n\) are:
\[(2,0,0),(0,2,0),(0,0,2)\]
\[(1,1,0),(1,0,1),(0,1,1)\]

So we have:

\[(a+b+c)^2={2 \choose 2, 0, 0}a^{2}b^{0}c^{0}+{2 \choose 0, 2, 0}a^{0}b^{2}c^{0}+{2 \choose 0, 0, 2}a^{0}b^{0}c^{2}\\
{2 \choose 1, 1, 0}a^{1}b^{1}c^{0}+{2 \choose 1, 0, 1}a^{1}b^{0}c^{1}+{2 \choose 0, 1, 1}a^{0}b^{1}c^{1}\]
Using the definition of the multinomial coefficient (all those big parentheses):
\[{2 \choose 2, 0, 0}={2 \choose 0, 2, 0}={2 \choose 0, 0, 2}=1\]
Similarly,
\[{2 \choose 1, 1, 0}={2 \choose 1, 0, 1}={2 \choose 0, 1, 1}=2\]
Thus,
\[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc\]

Now, here's the fun part. We can also use binomial theorem:
\[(u+v)^n=\sum_{0 \leq i \leq n}u^{n-i}v^{i}\]
Just let \[u=a+b \text{ and } v=c\]
Thus,
\[(a+b+c)^{n}=\sum_{0 \leq i \leq n}(a+b)^{n-i}c^i\]

There's a twist. We can apply binomial theorem again.

\[(a+b)^{q}=\sum_{0 \leq j \leq q}a^{q-j}b^{j}\]

So, let \[q=n-i\]
Thus,
\[(a+b+c)^{n}=\sum_{0 \leq i \leq n} \sum_{0 \leq j \leq n-i}a^{n-i-j}b^{j}c^{i}\]

(We could also show how the multinomial reduces to this, but it's not nearly as straightforward..)

Answer 27

Yeah, I understood all of that.