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How to simplify this best $(a_1 + a_2 +a_3+... +a_n)^m$ for $m=n, m<n, m>n$

I could only get $\sum_{i=0}^{m}\binom{m}{i}a_i^i\sum_{j=0}^{m-i}\binom{m-i}{j}a_j ... $

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It is already as simple as it gets if you don't have other information on the $a_i$s. –  Phira May 30 '12 at 15:06
    
they are just variable ... let's say like a and b in binomial expansion –  Santosh Linkha May 30 '12 at 15:07
    
Note that Pascal's Tetraedron, then its $4th$ dimensional analogue, then the $5th$, is a graphical way to visualise the multinomial theorem. –  Alyosha May 5 '13 at 16:59

2 Answers 2

up vote 8 down vote accepted

The simplification for this type of expansion is done through the multinomial theorem. The multinomial theorem is a generalization of the binomial case to any arbitrary number of terms in the sum to be exponentiated.

The multinomial theorem is written as follows:

$$(x_1 + x_2 + \cdots + x_m)^n = \sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m} \prod_{1\le t\le m}x_{t}^{k_{t}}\ $$

Where the multinomial co-efficient is defined as:

$$ {n \choose k_1, k_2, \ldots, k_m} = \frac{n!}{k_1!\, k_2! \cdots k_m!} $$

It may also be useful to you to note that the multinomial co-efficient is always expressible as products of binomial co-efficients [Graham, Knuth, Patashnik, Concrete Mathematics (2nd edition)]:

$$ {n \choose k_1, k_2, \ldots, k_m} = {x_1+x_2+\cdots+x_m \choose x_2+\cdots+x_m}\cdots{x_{m-1}+x_m \choose x_m} $$

A fuller explanation can be found on Wikipedia, or Wolfram MathWorld

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is there any relation between $ k_1, k_2, ... $ isn't $ k_1+k_2+ ...+k_m=n $ again going to be combination of k's?? –  Santosh Linkha May 30 '12 at 15:18
    
The only relation is that the sum of all $k_i$ is equal to n. I do not believe there is an equality relationship such as $k_1>k_2$ or anything like that. –  Shaktal May 30 '12 at 15:20
    
Oh ... thank you both!! –  Santosh Linkha May 30 '12 at 15:21

This is called the Multinomial theorem: $$(a_1 + a_2 +a_3+... +a_n)^m=\sum_{k_1+k_2+...+k_n=m}\frac{m!}{k_1!\cdot...\cdot k_n!}a_1^{k_1}\cdot...\cdot a_n^{k_n}$$

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