Coefficient of a polynomial defined by Legendre polynomial

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Homework Statement


The polynomial of order ##(l-1)## denoted ## W_{l-1}(x) ## is defined by
## W_{l-1}(x) = \sum_{m=1}^{l} \frac{1}{m} P_{m-1}(x) P_{l-m}(x) ## where ## P_m(x) ## is the Legendre polynomial of first kind. In addition, one can also write
## W_{l-1}(x) = \sum_{n=0}^{l-1} a_n \cdot x^n ##

Find the coefficient ## a_n ## in terms of ## n ## and ## l ##.

2. The attempt at a solution
I think the binomial form of ## P_m(x) ## would help
## P_m(x) = 2^m \cdot \sum_{k=0}^{m} C^{k}_{m} C^{\frac{m+k-1}{2}}_{m} x^k ##, with ## C^{k}_{m} = \frac{m!}{k!(m-k)!} ##. The next thing is to know "how to count" the number of terms in both expressions of ##W_{l-1}(x)##. This is where I stuck at.
 
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I've found the solution which is of the following form

## a_n = \sum_{m=1}^{l} \frac{1}{m} \sum_{i=0}^{n} a_i^{(m-1)} a_{n-i}^{(l-m)} ##

where a_i^(m-1) is the coefficient corresponding to the power ## x^i ## of the polynomial ## P_{m-1}(x) ## (the same convention for ## P_{l-m}(x) ##).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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