We have, $$ \left(1+x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{2 n} $$ Put $x=1$, we get $$ 3^n=a_0+a_1+a_2+a_3+\ldots+a_2 n $$ Put $x=i^2$, we get $$ 1=a_0+a_1+a_2+a_3+\ldots+a_{2 n} $$ Adding Eqs. (i) and (ii), we get $$ \begin{aligned} & 3^n+1=2\left(a_0+a_2+a_4+\ldots+a_{2 n}\right) \\ \therefore \quad & a_0+a_2+a_4+\ldots+a_{2 n}=\frac{1}{2}\left(3^n+1\right) \end{aligned} $$ Subtract Eq. (ii)/from Eq. (i) we get $$ \begin{aligned} & a_1+a_3+a_5+\ldots+a_{2 n-1}=\frac{1}{2}\left(3^n-1\right) \\ & \left(1+x+x^2\right)^n=a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{2 n} \end{aligned} $$ On differentiating both side with respect to $x$ we get $$ \begin{aligned} n\left(1+x+x^2\right)^{n-1}(1+2 x)=a_1 & +2 a_2 x+3 a_3 x^2 \\ & +\ldots+2 n a_{2 n} x^{2 n-1} \end{aligned} $$ put $x=1$ we get $$ \begin{array}{ll} & n \cdot 3^n=a_1+2 a_2+3 a_3+\ldots+2 n a_{2 n} \\ \therefore \quad & \mathrm{A} \rightarrow \mathrm{III}, \mathrm{B} \rightarrow \mathrm{IV}, \mathrm{C} \rightarrow \mathrm{II} \end{array} $$