Let n geq 3 be an integer. For a permutation | vedclass

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(A) For a polynomial $f_\sigma(x) = a_n x^{n-1} + a_{n-1} x^{n-2} + \ldots + a_1 = 0$,the sum of the roots $S_\sigma$ is given by Vieta's formulas as

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$S < -n!$

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(A) For a polynomial $f_\sigma(x) = a_n x^{n-1} + a_{n-1} x^{n-2} + \ldots + a_1 = 0$,the sum of the roots $S_\sigma$ is given by Vieta's formulas as $S_\sigma = -\frac{a_{n-1}}{a_n}$.The total sum $S$ is the sum over all $n!$ permutations of $(1, 2, \ldots, n)$ of the values $-\frac{a_{n-1}}{a_n}$.By symmetry,for any pair of distinct indices $i, j \in \{1, 2, \ldots, n\}$,the number of permutations where $a_n = i$ and $a_{n-1} = j$ is $(n-2)!$.Thus,$S = \sum_{\sigma} -\frac{a_{n-1}}{a_n} = -(n-2)! \sum_{i=1}^n \sum_{j 
eq i} \frac{j}{i}$.We can rewrite the inner sum as $\sum_{i=1}^n \frac{1}{i} ((\sum_{k=1}^n k) - i) = \sum_{i=1}^n \frac{1}{i} (\frac{n(n+1)}{2} - i) = \frac{n(n+1)}{2} \sum_{i=1}^n \frac{1}{i} - n$.Since $n \geq 3$,$\sum_{i=1}^n \frac{1}{i} > 1 + \frac{1}{2} + \frac{1}{3} = \frac{11}{6}$.Then $S = -(n-2)! [\frac{n(n+1)}{2} \sum_{i=1}^n \frac{1}{i} - n]$.For $n=3$,$S = -(1)! [\frac{3(4)}{2} (1 + \frac{1}{2} + \frac{1}{3}) - 3] = -[6(\frac{11}{6}) - 3] = -[11-3] = -8$. Since $n! = 6$,$S As $n$ increases,the magnitude of $S$ grows much faster than $n!$,thus $S < -n!$.

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