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As I am not good at math, I would like to construct an expression having the form like:

$ a^n (\sum_{i=0}^{n-1} \lambda_i \cdot b^i ) +(\sum_{i=0}^{n-1} \lambda_i \cdot a^i)\cdot b^n + \lambda_n\cdot a^n \cdot b^n, \quad (1)$

where $\lambda_i$ is a free parameter to be determined.

My goal is to determine $\lambda_i$ such that Express (1) has a neat and simple form. $\lambda_i \ (i=1,...,n)$ can be any real positive number, but they should satsify $\lambda_i < \lambda_{i+1}$ for $i=0,...,n-1$. For example, when $n=3$,

$ 1a^3 + 6a^3b+15a^3b^2+20a^3b^3+15a^2b^3+6ab^3+1b^3.$

Currently, I choose $\lambda_{i} = {2n\choose{i}}$ which is a bionormal coefficient. But it seems that Express (1) is hard to be simplied in this case.

My question is that can anyone give me a hand to find a good coefficient $\lambda_i$ to simply Express (1) in a neat form. Thanks!

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    Well, there's no way you're going to get a form *that* neat. What you have is not going to be a power of a binomial.2012-09-30

2 Answers 2

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You could take $\lambda_i={n\choose i}n^i$. Then your expression would be $(a+abn)^n+(b+abn)^n-(abn)^n$

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There is a nice formula for the partial sum of a geometric series:

$\sum_{i=0}^{n-1}{x^i} = \frac{1 - x^n}{1-x}$

So your expression will simplify if you let $\lambda_i = c^i$ for some choice of $c$. Since you want the $\lambda_i$'s to be increasing, you could choose $\lambda_i = 2^i$. Then you get:

$ a^n (\sum_{i=0}^{n-1} \lambda_i \cdot b^i ) +(\sum_{i=0}^{n-1} \lambda_i \cdot a^i)\cdot b^n + \lambda_n\cdot a^n \cdot b^n$

$= a^n (\sum_{i=0}^{n-1} 2^i \cdot b^i ) +(\sum_{i=0}^{n-1} 2^i \cdot a^i)\cdot b^n + 2^n\cdot a^n \cdot b^n$ $=a^n\frac{ 1-(2b)^n}{1-2b} +b^n\frac{ 1-(2a)^n}{1-2a} + (2ab)^n.$

That seems relatively simple to me - at least, it's closed form.