In expansion of (2x + 1).(2x + 5) . (2x + 9) . (2x + 13)...(2x + 49), coefficient of x^{12} is :-

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7.Binomial Theorem

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In the expansion of

$(2x + 1).(2x + 5) . (2x + 9) . (2x + 13)...(2x + 49),$ find the coefficient of $x^{12}$ is :-

A

$\frac{325}{2}$

B

$325 \cdot (2^{12})$

C

$325 \cdot (2^{11})$

D

None of these

Solution

These are $13$ factors so $2^{13}\left(\mathrm{x}+\frac{1}{2}\right)\left(\mathrm{x}+\frac{5}{2}\right)$

$\left(x+\frac{9}{2}\right) \cdots\left(x+\frac{49}{2}\right)$

$ \text { so coefficient of } x^{12} =2^{13}\left(\frac{1}{2}+\frac{5}{2}+\frac{9}{2}+\cdots+\frac{49}{2}\right) $

$=325.2^{12} $

Standard 11

Mathematics

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