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In a question in Fuhrmann's "A polynomial approach to linear algebra" it is stated that $$\sum_{i=1}^n \frac{p(a_i)}{q'(a_i)}=p_{a_{n-1}},$$ where $p,q$ are polynomials over a field with $deg(p)=n-1$ and $deg(q)=n$. $q(x)$ is also a monic polynomial with the unique factorization $p(x)=(x-a_1)(x-a_2)...(x-a_n)$.

$q'(x)$ is the formal derivative of $q(x)$, and $p_{a_{n-1}}$ is the $n-1$:th coefficient of $p$, i.e. it's leading coefficient.

Does anyone know how to do it?

A tip is to use Langrange interpolation.

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Use Lagrange interpolation for what? Pointer to where? Perhaps you forgot to state that you're interested in proving this statement? –  joriki Sep 28 '12 at 11:07
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By Lagrange interpolation, we have \[ p(x) = \sum_{i=1}^n p(a_i) \prod_{j \ne i} \frac{x - a_j}{a_i - a_j} \] As $q(x) = \prod_j (x-a_j)$, we have $q'(x) = \sum_i \prod_{j \ne i} (x-a_j)$, so $q'(a_i) = \prod_{j\ne i} (a_i-a_j)$, and $\prod_{j \ne i} (x-a_j)$ equals $\frac{q(x)}{x-a_i}$, so \[ p(x) = \sum_{i=1}^n \frac{p(a_i)}{q'(a_i)} \cdot \frac{q(x)}{(x-a_i)} \] Looking at the coefficient of $x^{n-1}$, gives, as $\frac{q(x)}{x-a_i}$ is monic \[ p_{n-1} = \sum_{i=1}^n \frac{p(a_i)}{q'(a_i)}. \]

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Thanks a lot for a very clear and helpful answer. Completely solved it for me. Have a nice weekend! –  user36773 Sep 28 '12 at 11:39
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