What is the coefficient of $x^{19}$ in the expansion of $ \prod \limits_{n=1}^{20} (x+n^2)$?

Question

How could we find the coefficient of $x^{19}$ in the expansion of $ \prod \limits_{n=1}^{20} (x+n^2)$?

Answer 1

Use Vieta's formula, noting that your polynomial has roots $x_k = -k^2$ for $k$ from 1 to 20.

Answer 2

$$ \prod \limits_{i=1}^d (x + a_i) = x^d + \Big(\sum_i a_i \Big) x^{d-1} + \Big( \sum_{i < j} a_i a_j \Big) x^2 + \Big( \sum_{i < j < k} a_i a_j a_k \Big) x^3 + \cdots, $$ so the coefficient of $x^{d-1}$ in $\prod \limits_{i=1}^d (x + a_i)$ is $\sum \limits_{i=1}^d a_i$. In your case, the answer is $\sum \limits_{i=1}^{20} i^2 = \cdots$.

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Note: If you are interested in the general term, the coefficient of $t^{\rm th}$ is given by the "elementary symmetric polynomial" $$ \sum_{i_1 < i_2 < \cdots < i_t} a_{i_1} a_{i_2} \cdots a_{i_t}. $$