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

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Question

(A) We have the expression $(1+x^2)^5(1+x)^4$. Using the binomial expansion formula $(1+a)^n = \sum_{k=0}^{n} {^nC_k} a^k$: $(1+x^2)^5 = {^5C_0} + {^5

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) We have the expression $(1+x^2)^5(1+x)^4$. Using the binomial expansion formula $(1+a)^n = \sum_{k=0}^{n} {^nC_k} a^k$: $(1+x^2)^5 = {^5C_0} + {^5C_1}x^2 + {^5C_2}x^4 + {^5C_3}x^6 + \dots = 1 + 5x^2 + 10x^4 + 10x^6 + \dots$ $(1+x)^4 = {^4C_0} + {^4C_1}x + {^4C_2}x^2 + {^4C_3}x^3 + {^4C_4}x^4 = 1 + 4x + 6x^2 + 4x^3 + x^4$ To find the coefficient of $x^5$,we multiply terms from the two expansions such that the sum of their powers is $5$: $(5x^2) \cdot (4x^3) + (10x^4) \cdot (4x) = 20x^5 + 40x^5 = 60x^5$ 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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