Let $P (x) = 24 x^{24} + \sum_{j = 1}^{23} (24 - j) (x^{24 - j} + x^{24 + j})$ . Let $x_{1}, x_{2} ..... x_{r}$ be the distinct zeros of P(x) and let ${x_{k}}^{2} = a_{k} + i b_{k}$ for k = 1 2, ..... r where $a_{k}, b_{k}$ are real numbers, and $i^{2} = - 1$ . Let $\sum_{k = 1}^{r} | b_{k} | = m + n \sqrt{p}$ where m, n and p are integers and p is prime then the value of $m + n + p$ is equal to _____

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answer is D.

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Detailed Solution

$P (x) = 24 x^{24} + 23 (x^{23} + x^{25}) + 22 (x^{22} + x^{26}) + ..... + 1 (x^{13} + x^{47}) = x + 2 x^{2} + ........ + 24 x^{24} + ...... + x^{47} = x {(1 + x + ...... + x^{23})}^{2} = \frac{x {(x^{24} - 1)}^{2}}{{(x - 1)}^{2}} P (x) = 0 \Rightarrow x^{24} = 1 x = 0, α, α^{2} ....... α^{23} w h e r e α = e^{i 2 π / 24} \sum_{K = 1}^{r} | b_{K} | = \sum_{K = 1}^{r} Im (| {x_{K}}^{2} |) = 8 + 4 \sqrt{3}$