In the expansion of (2x + 1)(2x + 5)(2x + 9)( | vedclass

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(B) The given expression is $P(x) = (2x + 1)(2x + 5)(2x + 9) \cdots (2x + 49)$.This is a product of $13$ terms of the form $(2x + a_k)$,where $a_k = 1

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$325 \cdot 2^{12}$

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(B) The given expression is $P(x) = (2x + 1)(2x + 5)(2x + 9) \cdots (2x + 49)$.This is a product of $13$ terms of the form $(2x + a_k)$,where $a_k = 1 + (k-1)4$ for $k = 1, 2, \dots, 13$.We can factor out $2$ from each of the $13$ terms:$P(x) = 2^{13} \left(x + \frac{1}{2}ight) \left(x + \frac{5}{2}ight) \left(x + \frac{9}{2}ight) \cdots \left(x + \frac{49}{2}ight)$.The coefficient of $x^{12}$ in the expansion of $(x + a_1)(x + a_2) \cdots (x + a_n)$ is the sum of the roots taken $(n-1)$ at a time,which is $\sum_{i=1}^{n} a_i$ if we consider the expansion as $x^n + (\sum a_i)x^{n-1} + \dots$. Wait,for $n=13$,the coefficient of $x^{12}$ is the sum of the constants: $S = \sum_{k=1}^{13} \frac{a_k}{2}$.$S = \frac{1}{2} + \frac{5}{2} + \frac{9}{2} + \dots + \frac{49}{2} = \frac{1}{2} (1 + 5 + 9 + \dots + 49)$.This is an arithmetic progression with $n=13$,first term $a=1$,last term $l=49$.Sum $= \frac{13}{2} (1 + 49) = \frac{13 	imes 50}{2} = 325$.Thus,the coefficient of $x^{12}$ is $2^{13} 	imes \left(\frac{325}{2}ight) = 325 	imes 2^{12}$.

In the expansion of $(2x + 1)(2x + 5)(2x + 9)(2x + 13) \cdots (2x + 49)$,find the coefficient of $x^{12}$.

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