The coefficient of x^5 in the expansion of (1 | vedclass

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(B) We have $(1 + x^2)^5(1 + x)^4$.Using the binomial theorem,the expansion is:$(^5C_0 + ^5C_1x^2 + ^5C_2x^4 + ^5C_3x^6 + ...)(^4C_0 + ^4C_1x + ^4C_2x

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) We have $(1 + x^2)^5(1 + x)^4$.Using the binomial theorem,the expansion is:$(^5C_0 + ^5C_1x^2 + ^5C_2x^4 + ^5C_3x^6 + ...)(^4C_0 + ^4C_1x + ^4C_2x^2 + ^4C_3x^3 + ^4C_4x^4)$.To find the coefficient of $x^5$,we look for terms whose product results in $x^5$:$1$. The term $^5C_2x^4$ from the first bracket multiplied by $^4C_1x$ from the second bracket gives $^5C_2 	imes ^4C_1 	imes x^5 = 10 	imes 4 	imes x^5 = 40x^5$.$2$. The term $^5C_1x^2$ from the first bracket multiplied by $^4C_3x^3$ from the second bracket gives $^5C_1 	imes ^4C_3 	imes x^5 = 5 	imes 4 	imes x^5 = 20x^5$.Adding these coefficients,we get $40 + 20 = 60$.Thus,the coefficient of $x^5$ is $60$.

The coefficient of $x^5$ in the expansion of $(1 + x^2)^5(1 + x)^4$ is

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