Find the coefficient of a^{4} in the product | vedclass

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Question

(A) We first expand each of the factors of the given product using the Binomial Theorem. We have $(1 + 2a)^4 = {^4C_0} + {^4C_1}(2a) + {^4C_2}(2a)^2 +

Vedclass · Accepted Answer

$-438$

Vedclass · Answer

(A) We first expand each of the factors of the given product using the Binomial Theorem. We have $(1 + 2a)^4 = {^4C_0} + {^4C_1}(2a) + {^4C_2}(2a)^2 + {^4C_3}(2a)^3 + {^4C_4}(2a)^4 = 1 + 8a + 24a^2 + 32a^3 + 16a^4$. And $(2 - a)^5 = {^5C_0}(2)^5 - {^5C_1}(2)^4(a) + {^5C_2}(2)^3(a)^2 - {^5C_3}(2)^2(a)^3 + {^5C_4}(2)(a)^4 - {^5C_5}(a)^5 = 32 - 80a + 80a^2 - 40a^3 + 10a^4 - a^5$. To find the coefficient of $a^4$ in the product $(1 + 8a + 24a^2 + 32a^3 + 16a^4)(32 - 80a + 80a^2 - 40a^3 + 10a^4 - a^5)$,we multiply terms that result in $a^4$: $1(10a^4) + (8a)(-40a^3) + (24a^2)(80a^2) + (32a^3)(-80a) + (16a^4)(32) = 10a^4 - 320a^4 + 1920a^4 - 2560a^4 + 512a^4 = -438a^4$. Thus,the coefficient of $a^4$ is $-438$.

Find the coefficient of $a^{4}$ in the product $(1+2a)^{4}(2-a)^{5}$ using the binomial theorem.

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