The coefficient of x^{12} in the expansion of | vedclass

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(C) We need to find the coefficient of $x^{12}$ in $(x^2+2x+2)^8$. Let $f(x) = (x^2+2x+2)^8 = ((x+1)^2+1)^8$. Using the binomial expansion,$((x+1)^2+1

Vedclass · Accepted Answer

$2576$

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(C) We need to find the coefficient of $x^{12}$ in $(x^2+2x+2)^8$. Let $f(x) = (x^2+2x+2)^8 = ((x+1)^2+1)^8$. Using the binomial expansion,$((x+1)^2+1)^8 = \sum_{k=0}^{8} \binom{8}{k} (x+1)^{2k}$. We want the coefficient of $x^{12}$. The term $(x+1)^{2k}$ contains $x^{12}$ if $2k \ge 12$,i.e.,$k \ge 6$. The general term in the expansion is $\binom{8}{k} \binom{2k}{12} x^{12}$. Summing over $k=6, 7, 8$: For $k=6$: $\binom{8}{6} \binom{12}{12} = 28 	imes 1 = 28$. For $k=7$: $\binom{8}{7} \binom{14}{12} = 8 	imes \binom{14}{2} = 8 	imes 91 = 728$. For $k=8$: $\binom{8}{8} \binom{16}{12} = 1 	imes \binom{16}{4} = 1 	imes 1820 = 1820$. Total coefficient = $28 + 728 + 1820 = 2576$.

The coefficient of $x^{12}$ in the expansion of $(x^2+2x+2)^8$ is

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