The coefficient of x^4 in the expansion of (1 | vedclass

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(D) We have $(1 + x + x^2 + x^3)^n = ((1 + x)(1 + x^2))^n = (1 + x)^n (1 + x^2)^n$. Expanding both parts using the Binomial Theorem: $(1 + x)^n = 1 +

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$^nC_4 + ^nC_2 + ^nC_1 \cdot ^nC_2$

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(D) We have $(1 + x + x^2 + x^3)^n = ((1 + x)(1 + x^2))^n = (1 + x)^n (1 + x^2)^n$. Expanding both parts using the Binomial Theorem: $(1 + x)^n = 1 + ^nC_1 x + ^nC_2 x^2 + ^nC_3 x^3 + ^nC_4 x^4 + \dots$ $(1 + x^2)^n = 1 + ^nC_1 x^2 + ^nC_2 x^4 + \dots$ To find the coefficient of $x^4$,we multiply terms from both expansions such that the sum of powers is $4$: $1 \cdot (^nC_2 x^4) + (^nC_2 x^2) \cdot (^nC_1 x^2) + (^nC_4 x^4) \cdot 1$ $= ^nC_2 x^4 + ^nC_1 \cdot ^nC_2 x^4 + ^nC_4 x^4$ $= (^nC_4 + ^nC_2 + ^nC_1 \cdot ^nC_2) x^4$. Thus,the coefficient is $^nC_4 + ^nC_2 + ^nC_1 \cdot ^nC_2$.

The coefficient of $x^4$ in the expansion of $(1 + x + x^2 + x^3)^n$ is

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