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Let $P(x) := \prod_{k = 1}^{50} (x - k)$ and $Q(x) := \prod_{k = 1}^{50} (x + k)$. If $P(x) Q(x) = \sum_{k = 0}^{100} a_k x^k$, find $a_{100} - a_{99} - a_{98} - a_{97}$.

The correct answer is 42926. Can you explain the solution to this problem?

I have used the formula for the sum of roots for $P(x)Q(x)$ (Vieta's formula), and concluded $a_{99}=0$. $$ \sum_{k = 1}^{n} r_k = - \frac{a_{n - 1}}{a_n} $$

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  • $\begingroup$ I tried to apply the formula for the sum of roots and concluded that a99=0 $\endgroup$ – Vasanth Bhaskara Jun 13 at 11:09
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    $\begingroup$ I haven't downvoted, but I believe I know why it has been done: Your post currently doesn't comply with our guidelines for good questions. In particular, you have not provided what we call context. You have said some useful things in the comments here, but it would be best if that was gathered in the actual post. Also, we generally frown upon using images of text rather than actual text. $\endgroup$ – Arthur Jun 13 at 11:22
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Let $$Q(x)=\sum_{i=1}^{50} b_ix^i.$$ Then $P(x)=Q(-x)$, so $$P(x)=\sum_{i=1}^{50} (-1)^ib_ix^i.$$

$$a_{100}=b_{50}^2=1$$ $$a_{99}=b_{50}b_{49}-b_{49}b_{50}=0$$ $$a_{98}=2b_{48}b_{50}-b_{49}^2$$ $$=\sum_{i\ne j} ij - \left(\sum_i i\right)^2=-\sum_{i=1}^{50} i^2=-42925.$$ $$a_{97}=b_{50}b_{47}-b_{49}b_{48}+b_{48}b_{49}-b_{47}b_{50}=0.$$

Thus we get $1-(-42925)=42926$ as an answer.

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My approach would be to do a quick inspection: $$ P(x)Q(x) =\prod_{i=1}^{50} (x+i)(x-i) = \prod_{i=1}^{50} (x^2-i^2). $$ This immediate tells us two things:

1) We only need to worry about even coefficients

2) Everything will be terms of squares

Consider $a_{100}$ it must be $1$ since the only way to get to $x^{100}$ from this requires multiplying the $x$ terms together.

For $a_{98}$ we pick one term out of the product to be the non-x term and the rest are the $x$ to ensure that we get a power of 97. This gives $\sum_i i^2 = \frac{n(n+1)(2n+1)}{6} = 42925$ so $1-(-42925) = 42926$

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  • $\begingroup$ Can you please explain what you mean by (2) and "one out of the product to be the $i$ and the rest are the $x$"? $\endgroup$ – Viktor Glombik Jun 16 at 9:06
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We know that the roots of $P$ are $\{1, \ldots, 50\}$ and the roots of $Q$ are $\{-1, \ldots, -50\}$. Therefore the roots of $P \cdot Q$ are $\{-50, \ldots, -2, -1, 1, 2, \ldots, 50\}$, which we'll call $r_1, \ldots, r_{100}$. From the first of Vietas formulas we know that $$ 0 = -50 + \ldots + -2 + -1 + 1 + 2 + \ldots 50 = \sum_{k = 0}^{100} r_{k} = - \frac{a_{99}}{a_{100}}. $$ This shows $a_{99} = 0$.

From the second formula we see that \begin{align} -42925 & = \sum_{k = 1}^{50} (- k^2) -50(-49 + 48 + \ldots + -1 + 1 + \ldots + 49 + 50) + (-49)(50 + 49) + \ldots + (49)(50) \\ & = -50^2 + (-49)(49 + 50) + (-48)(48 + 49 + 50) + \ldots + (49)(50)\\ & = -50^2 - 49^2 - 48^2 - \ldots - 1 \\ & = \sum_{k = 1}^{100} \sum_{\ell = k}^{100} r_k r_{\ell} = \frac{a_{98}}{a_{100}}, \end{align}

From the third formula we get that $$ \prod_{k = 1}^{50} k^2 = (-50)(50) \ldots (-2)(2)(-1)(1) = \prod_{k = 1}^{100} r_k = (-1)^{100} \frac{a_0}{a_{100}} = \frac{a_0}{a_{100}}, $$ Lastly we have $a_0 = \prod_{k = 1}^{100} r_k$ by easy inspection. Then we have $a_{100} = \frac{a_{0}}{\prod_{k = 1}^{100} r_k} = 1$ from the third formula and therefore $a_{98} = \sum_{k = 1}^{5} (-k^2) = -42925$ yielding $$ a_{100} - a_{99} - a_{98} - a_{97} = 1 - 0 - (-42925) - a_{97} = 42926 - a_{97}. $$ Now since $P(x)Q(x) = \prod_{k = 1}^{50} (x^k - k^2)$ only the even coefficients are non-zero. This proves the claim.

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  • $\begingroup$ I think you have it here Viktor! $\endgroup$ – Kitter Catter Jun 15 at 17:36
  • $\begingroup$ @KitterCatter I edited after seeing the new answer, realising I had only made a small error as I expected. $\endgroup$ – Viktor Glombik Jun 15 at 18:19

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