IIT JEE , AIEEE Forum - Mathematics , Algebra

neeraj

I think you know the binomial theorem I would start with trinomials. You should expand (a + b + c)² and (a + b + c)³ by multiplying them out to see how the coefficients are constructed.
(a + b + c)²
=(a + b + c)(a + b + c)
= a² + b² + c² + 2ab + 2ac + 2bc

(a + b + c)³
= (a + b + c)(a + b + c)(a + b + c)
= a³ + b³ + c³ + 3a²b + 3a²c + 3ab² + 3b²c + 3 ac² + 3 bc² + 6 abc

If you think of these two examples as (a + b + c)ⁿ then each of the term is a constant times a^pb^qc^r where p + q + r = n. In the first case n = 2 and in the second n = 3. In the general case

The notation means that you form a sum of all possible terms of the form a^pb^qc^r where p + q + r = n with each term having a coefficient of

The multinomial theorem is of the same form.

Multinomial Theorem Proof (Alternate)
A multinomial is a mathematical expression consisting of two or more terms, e.g.

$\displaystyle a_1 x_1 + a_2 x_2 + \ldots + a_k x_k.$

The multinomial theorem provides the general form of the expansion of the powers of this expression, in the process specifying the multinomial coefficients which are found in that expansion. The expansion is:

$\displaystyle (x_1 + x_2 + \ldots + x_k)^n = \sum \frac{n!}{n_1! n_2! \dotsb n_k!} x_1^{n_1} x_2^{n_2} \cdots x_k^{n_k}$ (1)

where the sum is taken over all multi-indices $(n_1, \ldots n_k)\in\mathbb{N}^k$ that sum to .

The expression $\frac{n!}{n_1! n_2! \cdots n_k!}$ occurring in the expansion is called multinomial coefficient and is denoted by

$\displaystyle \binom{n}{n_1, n_2, \dotsc, n_k}.$
Proof. The below proof of the multinomial theorem uses the binomial theorem and induction on

. In addition, we shall use multi-index notation.

First, for

, both sides equal

. For the induction step, suppose the multinomial theorem holds for

. Then the binomial theorem and the induction assumption yield

$\displaystyle (x_1+\cdots + x_k\,+\,x_{k+1})^n$	$\displaystyle =$	$\displaystyle \sum_{l=0}^n {n \choose l} (x_1+\cdots + x_k)^l x_{k+1}^{n-l}$
	$\displaystyle =$	$\displaystyle \sum_{l=0}^n {n \choose l} l! \sum_{\vert i\vert=l} \frac{x^i}{i!} x_{k+1}^{n-l}$
	$\displaystyle =$	$\displaystyle n! \sum_{l=0}^n \sum_{\vert i\vert=l} \frac{x^i x_{k+1}^{n-l}}{i! (n-l)!}$

where $x=(x_1,\ldots, x_k)$ and

is a multi-index in

. To complete the proof, we need to show that the sets

$\displaystyle A$	$\displaystyle =$	$\displaystyle \{ (i_1,\ldots,i_k, n-l)\in I^{k+1}_+ \mid l=0,\ldots, n,\, \vert(i_1,\ldots, i_k)\vert=l \},$
$\displaystyle B$	$\displaystyle =$	$\displaystyle \{j \in I^{k+1}_+ \mid \vert j\vert=n \}$

are equal. The inclusion $A \subset B$ is clear since

$\displaystyle \vert(i_1,\ldots,i_k, n-l)\vert = l + n-l = n.$

For $B \subset A$ , suppose $j=(j_1,\ldots, j_{k+1}) \in I^{k+1}_+$ , and $\vert j\vert=n$ . Let $l=\vert(j_1,\ldots, j_k)\vert$ . Then $l=n-j_{k+1}$ , so $j_{k+1} = n-l$ for some $l=0,\ldots, n$ . It follows that that

.

Let us define $y=(x_1,\cdots, x_{k+1})$ and let $j=(j_1,\ldots, j_{k+1})$ be a multi-index in $I_+^{k+1}$ . Then

$\displaystyle (x_1+\cdots + x_{k+1})^n$	$\displaystyle =$	$\displaystyle n! \sum_{\vert j\vert=n} \frac{x^{(j_1,\ldots, j_k)} x_{k+1}^{j_{k+1}}}{(j_1,\ldots, j_k)! j_{k+1}!}$
	$\displaystyle =$	$\displaystyle n! \sum_{\vert j\vert=n} \frac{y^j}{j!}.$

This completes the proof.

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