How to simplify this best $(a_1 + a_2 +a_3+... +a_n)^m$ for $m=n, m
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 ... $
How to simplify this best $(a_1 + a_2 +a_3+... +a_n)^m$ for $m=n, m
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 ... $
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
This is called the Multinomial theorem: $(a_1 + a_2 +a_3+... +a_n)^m=\sum_{k_1+k_2+...+k_n=m}\frac{m!}{k_1!\cdot...\cdot k_n!}a_1^{k_1}\cdot...\cdot a_n^{k_n}$