Let p(x) = $$n_0x^{2k+1}+n_0x^{2k}+n_1x^{2k-1}+n_1x^{2k-2}+n_2x^{2k-3}+n_2x^{2k-4}+.........+n_{k+1}x+n_{k+1}$$
where k is a positive integer and the coefficients of every two adjacent terms in the polynomial are equal. It is also important to note here that the polynomial has {2k+2} terms. (one term for each power of x from 0 - 2k+1)
The question is as follows:
If p(x) has roots $$\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5,\alpha_6,.........,\alpha_{2k},\alpha_{2k+1}$$
find the value of:
$$\tan^{-1}(\alpha_1)+\tan^{-1}(\alpha_2)+\tan^{-1}(\alpha_3)+\tan^{-1}(\alpha_4)+.........+\tan^{-1}(\alpha_{2k})+\tan^{-1}(\alpha_{2k+1})$$
ie. Find the value of the sum of all individual inverse tangents of the roots of p(x)
What I have found out so far:
If we take a polynomial of degree 3 that has roots $\alpha,\beta,\gamma$ $$f(x)=ax^3+bx^2+cx+d$$
then the value of $\tan^{-1}(\alpha)+\tan^{-1}(\beta)+\tan^{-1}(\gamma)$ can be worked out fairly easily:
$\tan^{-1}(\alpha)+\tan^{-1}(\beta)+\tan^{-1}(\gamma)$ $$=\tan^{-1}\left(\frac{\alpha+\beta}{1-\alpha\beta}\right)+\tan^{-1}(\gamma)$$
$$=\tan^{-1}\left(\frac{(\frac{\alpha+\beta}{1-\alpha\beta})+\gamma}{1-(\frac{\alpha+\beta}{1-\alpha\beta})(\gamma)}\right)$$
$$=\tan^{-1}\left(\frac{\alpha+\beta+\gamma-\alpha\beta\gamma}{1-(\alpha\beta+\alpha\gamma+\beta\gamma)}\right)$$
Which will obviously give us the sum, product and sum of products of the roots, leaving us with a nice answer in terms of only the coefficients of p(x).
However, I am unsure how I can generalise this in continuing onwards to higher degree polynomials? And why would the repeating coefficients or odd degree in p(x) change our answer?
I would very much appreciate any help I can get with this question.