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  4. Coefficient of x13 in the expansion of (1-x)5(1+x+x2+x3)4 is

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Q. Coefficient of $x^{13}$ in the expansion of $(1-x)^{5}\left(1+x+x^{2}+x^{3}\right)^{4}$ is

Binomial Theorem

Solution:

Expression $=(1-x)^{5} \cdot(1+x)^{4}\left(1+x^{2}\right)^{4}$
$=(1-x)\left(1-x^{2}\right)^{4}\left(1+x^{2}\right)^{4}$
$=(1-x)\left(1-x^{4}\right)^{4}$
$\therefore $ Coefficient of $x^{13}=-{ }^{4} C_{3}(-1)^{3}=4$