If $C_0, C_1, C_2, \ldots, C_{10}$ represent the binomial coefficients in the expansion of $(1+x)^{10}$,then $C_0 C_6+C_1 C_7+C_2 C_8+C_3 C_9+C_4 C_{10}=$

$3 \cdot C_0 + 7 \cdot C_1 + 11 \cdot C_2 + \ldots + (3 + 4n) C_n =$

The value of ${ }^{10} C_{1}+{ }^{10} C_{2}+{ }^{10} C_{3}+\ldots+{ }^{10} C_{9}$ is

If the coefficients of $a^{r-1}$,$a^{r}$,and $a^{r+1}$ in the expansion of $(1+a)^{n}$ are in arithmetic progression,prove that $n^{2}-n(4r+1)+4r^{2}-2=0$.

For $z \in \mathbb{C}$,if $(1+z)^n = 1 + { }^n C_1 z + { }^n C_2 z^2 + \ldots + { }^n C_n z^n$ and $\sum_{r=0}^{100} { }^{100} C_r \sin(rx) = \left(2 \cos \frac{x}{2}\right)^{100} \sin(kx)$,then $k =$

Match the expressions in List-$I$ with their values in List-$II$ for the expansion $(1+x+x^2)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n}$.

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

The correct match is: