Match the expressions in List-I with their va | vedclass

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(D) Given the expansion: $(1+x+x^2)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n}$.Step $1$: Put $x=1$:$(1+1+1)^n = a_0 + a_1 + a_2 + \ldots + a_

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$(d) A-III, B-IV, C-II$

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(D) Given the expansion: $(1+x+x^2)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n}$.Step $1$: Put $x=1$:$(1+1+1)^n = a_0 + a_1 + a_2 + \ldots + a_{2n} \implies 3^n = a_0 + a_1 + a_2 + \ldots + a_{2n} \quad (i)$Step $2$: Put $x=-1$:$(1-1+1)^n = a_0 - a_1 + a_2 - a_3 + \ldots + a_{2n} \implies 1 = a_0 - a_1 + a_2 - a_3 + \ldots + a_{2n} \quad (ii)$Step $3$: Adding $(i)$ and $(ii)$:$3^n + 1 = 2(a_0 + a_2 + a_4 + \ldots) \implies a_0 + a_2 + \ldots = \frac{1}{2}(3^n + 1)$. Thus,$A \rightarrow III$.Step $4$: Subtracting $(ii)$ from $(i)$:$3^n - 1 = 2(a_1 + a_3 + a_5 + \ldots) \implies a_1 + a_3 + \ldots = \frac{1}{2}(3^n - 1)$. Thus,$B \rightarrow IV$.Step $5$: Differentiating the expansion with respect to $x$:$n(1+x+x^2)^{n-1}(1+2x) = a_1 + 2a_2 x + 3a_3 x^2 + \ldots + 2n a_{2n} x^{2n-1}$.Put $x=1$:$n(3)^{n-1}(3) = a_1 + 2a_2 + 3a_3 + \ldots + 2n a_{2n} \implies n \cdot 3^n = a_1 + 2a_2 + 3a_3 + \ldots + 2n a_{2n}$. Thus,$C \rightarrow II$.Therefore,the correct match is $A-III, B-IV, C-II$.

List-$I$	List-$II$
$(A)$ $a_0 + a_2 + \ldots + a_{2n}$	$(I)$ $n \cdot 3^{n-1}$
$(B)$ $a_1 + a_3 + \ldots + a_{2n-1}$	$(II)$ $n \cdot 3^n$
$(C)$ $a_1 + 2a_2 + 3a_3 + \ldots + 2n a_{2n}$	$(III)$ $\frac{1}{2}(3^n + 1)$
	$(IV)$ $\frac{1}{2}(3^n - 1)$

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