Let $(1+2x)^{20} = a_0 + a_1x + a_2x^2 + \dots + a_{20}x^{20}$. Then $3a_0 + 2a_1 + 3a_2 + 2a_3 + 3a_4 + 2a_5 + \dots + 2a_{19} + 3a_{20}$ equals

Let $\alpha = \sum_{k=0}^n \left( \frac{({ }^n C_k)^2}{k+1} \right)$ and $\beta = \sum_{k=0}^{n-1} \left( \frac{{ }^n C_k \cdot { }^n C_{k+1}}{k+2} \right)$. If $5 \alpha = 6 \beta$,then $n$ equals:

The value of $\sum_{r=0}^{6} \left({}^{6}C_{r} \cdot {}^{6}C_{6-r}\right)$ is equal to:

Let $n \in N$ and $[x]$ denote the greatest integer less than or equal to $x$. If the sum of $(n+1)$ terms ${}^{n}C_{0}, 3 \cdot {}^{n}C_{1}, 5 \cdot {}^{n}C_{2}, 7 \cdot {}^{n}C_{3}, \ldots$ is equal to $2^{100} \cdot 101$,then $2\left[\frac{n-1}{2}\right]$ is equal to $....$

If $C_r = ^{100}C_r$,then the value of $1 \cdot C_0^2 - 2 \cdot C_1^2 + 3 \cdot C_2^2 - 4 \cdot C_3^2 + \dots + 101 \cdot C_{100}^2$ is equal to:

If $(1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n$,then $C_0 + 2 C_1 + 3 C_2 + \ldots + (n+1) C_n$ is equal to