Find the coefficient of x^{49} in the expansi | vedclass

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(B) The given expression is $P(x) = (2x + 1)(2x + 3)(2x + 5) \dots (2x + 99)$.Factor out $2$ from each of the $50$ terms: $P(x) = 2^{50} \left(x + \fr

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$2^{49} 	imes 2500$

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(B) The given expression is $P(x) = (2x + 1)(2x + 3)(2x + 5) \dots (2x + 99)$.Factor out $2$ from each of the $50$ terms: $P(x) = 2^{50} \left(x + \frac{1}{2}ight) \left(x + \frac{3}{2}ight) \dots \left(x + \frac{99}{2}ight)$.Let $a_k = \frac{2k-1}{2}$ for $k = 1, 2, \dots, 50$. Then $P(x) = 2^{50} \prod_{k=1}^{50} (x + a_k)$.The coefficient of $x^{49}$ in the expansion of $\prod_{k=1}^{50} (x + a_k)$ is the sum of the roots taken $49$ at a time,which is equivalent to the sum of all roots multiplied by $(-1)$ if we consider the Vieta's formula for a polynomial of degree $50$. Specifically,the coefficient of $x^{n-1}$ in $\prod (x+a_k)$ is $\sum a_k$.Thus,the coefficient of $x^{49}$ is $2^{50} 	imes \sum_{k=1}^{50} \frac{2k-1}{2} = 2^{49} 	imes \sum_{k=1}^{50} (2k-1)$.The sum of the first $50$ odd numbers is $50^2 = 2500$.Therefore,the coefficient is $2^{49} 	imes 2500$.

Find the coefficient of $x^{49}$ in the expansion of $(2x + 1)(2x + 3)(2x + 5) \dots (2x + 99)$.

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