# how to determine the coefficient to simplify this expression

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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You can get parentheses (or any other paired delimiters like brackets, braces, ...) of the appropriate size by preceding them with \left and ˚\right`. –  joriki Sep 28 '12 at 18:16
Is there a reason why the $a^nb^n$ term is written separately instead of extending the second sum to $n$? –  joriki Sep 28 '12 at 18:17
What would you consider to be a "neat form"? –  Gerry Myerson Sep 29 '12 at 6:34
For example, $(a+b)^3$ is a neat form of $a^3+3a^2b+3ab^2+b^3$. –  John Smith Sep 29 '12 at 8:15
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. –  Gerry Myerson Sep 30 '12 at 3:50

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.

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