The coefficient of x^8 in the expansion of (1 | vedclass

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(C) The expression is $(1 - x^4)^4 (1 + x)^5$. Using the Binomial Theorem,$(1 - x^4)^4 = \sum_{k=0}^{4} \binom{4}{k} (-x^4)^k = \binom{4}{0} - \binom{

Vedclass · Accepted Answer

$-14$

Vedclass · Answer

(C) The expression is $(1 - x^4)^4 (1 + x)^5$. Using the Binomial Theorem,$(1 - x^4)^4 = \sum_{k=0}^{4} \binom{4}{k} (-x^4)^k = \binom{4}{0} - \binom{4}{1}x^4 + \binom{4}{2}x^8 - \dots$ And $(1 + x)^5 = \sum_{r=0}^{5} \binom{5}{r} x^r = \binom{5}{0} + \binom{5}{1}x + \dots + \binom{5}{4}x^4 + \dots + \binom{5}{8}x^8$ (where $\binom{5}{8}=0$). We need the coefficient of $x^8$ in the product: $(\binom{4}{0} - \binom{4}{1}x^4 + \binom{4}{2}x^8) (\binom{5}{0} + \binom{5}{1}x + \dots + \binom{5}{4}x^4 + \dots + \binom{5}{8}x^8)$ The terms contributing to $x^8$ are: $1. \binom{4}{2}x^8 	imes \binom{5}{0} = 6 	imes 1 = 6$ $2. -\binom{4}{1}x^4 	imes \binom{5}{4}x^4 = -4 	imes 5 = -20$ $3. \binom{4}{0} 	imes \binom{5}{8}x^8 = 1 	imes 0 = 0$ Total coefficient $= 6 - 20 = -14$.

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

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