Coefficient of x^8 in expansion of (1 -x^4)^4 (1 + x)^5 is :-

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7.Binomial Theorem

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The coefficient of $x^8$ in the expansion of $(1 -x^4)^4 (1 + x)^5$ is :-

A

$20$

B

$-32$

C

$-14$

D

$30$

Solution

${x^4}$	${x^1}$
$4$	$4$
$8$	$0$

So coeff. of $x^{8}$ is

$ = { – ^4}{{\rm{C}}_1} \times {\,^5}{{\rm{C}}_4} + {\,^4}{{\rm{C}}_2} \times {\,^5}{{\rm{C}}_0}$

$=-20+6=-14$

Standard 11

Mathematics

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