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

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Question

$(2 n+1)(2 n+3)(2 n+5) \dots(2 x+99)$

$\Rightarrow 2^{50}\left[\left(\mathrm{x}+\frac{1}{2}\right)\left(\mathrm{x}+\frac{3}{2}\right)\left(\mathrm{

Vedclass · Accepted Answer

${2^{49}} 	imes 2500$

Vedclass · Answer

$(2 n+1)(2 n+3)(2 n+5) \dots(2 x+99)$

$\Rightarrow 2^{50}\left[\left(\mathrm{x}+\frac{1}{2}ight)\left(\mathrm{x}+\frac{3}{2}ight)\left(\mathrm{x}+\frac{5}{2}ight) \ldots\left(\mathrm{x}+\frac{99}{2}ight)ight]$

Coefficient of $\mathrm{x}^{49}$ is equal to $-(\mathrm{sum} 	ext { of roots })$

$=2^{50}\left[\frac{1}{2}+\frac{3}{2}+\frac{5}{2}+\ldots .+\frac{99}{2}ight]=2^{49} 	imes 2500$

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

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