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

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Try to answer the question for $20$ and $19$ replaced by something much smaller, like $3$ and $2$. Then try $4$ and $3$ and then $5$ and $4$. Then make a guess... –  t.b. Nov 22 '11 at 16:19
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2 Answers

up vote 2 down vote accepted

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

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$$ \prod \limits_{i=1}^d (x + a_i) = x^d + \Big(\sum_i a_i \Big) x^{d-1} + \Big( \sum_{i < j} a_i a_j \Big) x^2 + \Big( \sum_{i < j < k} a_i a_j a_k \Big) x^3 + \cdots, $$ 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 $\sum \limits_{i=1}^{20} i^2 = \cdots$.


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

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Could you please add few more terms of $\prod \limits_{i=1}^d (x + a_i) = x^d + \Big(\sum \limits_{i=1}^d a_i \Big) x^{d-1} + \Big( \sum \limits_{i < j} a_i a_j \Big) x^2 + \cdots,$ ? I am not able to spot the pattern. –  Quixotic Nov 22 '11 at 16:39
    
Also, how exactly $\Big( \sum \limits_{i < j} a_i a_j \Big)$ working here ? –  Quixotic Nov 22 '11 at 16:41
    
@MaX: roughly, take all $k$-subsets of the $a_k$, multiply all the members of those subsets, and then add up those products. –  J. M. Nov 22 '11 at 16:49
    
@MaX You have to take pairs of numbers, like: {a_1, a_2}, {a_1, a_3}, {a_1, a_4}, ...,{a_1, a_n}, {a_2, a_3}, {a_2, a_4}, ..., {a_2, a_n}, ..., {a_{n-1}, a_n}. Multiply each pair and all those up to get a big sum. For a general k-th term, you should generalize this as J.M. describes. –  Srivatsan Nov 22 '11 at 16:51
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