If $^nC_r = C_r$ and $2 \frac{C_1}{C_0} + 4 \frac{C_2}{C_1} + 6 \frac{C_3}{C_2} + \dots + 2n \frac{C_n}{C_{n-1}} = 650$,then $^nC_2 =$

If $C_j = {}^{n}C_j$,then $C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \ldots + C_{n-r} C_n = $

Let $S_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$,$S_2 = \sum_{j=1}^{10} j \binom{10}{j}$,and $S_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$.
Assertion $(A) : S_3 = 55 \times 2^9$
Reason $(R) : S_1 = 90 \times 2^8$ and $S_2 = 10 \times 2^8$

The ratio of the coefficients of the terms $x^{n-r}a^r$ and $x^ra^{n-r}$ in the binomial expansion of $(x+a)^n$ is:

If $(1+x-2x^2)^6 = 1+a_1x+a_2x^2+\ldots+a_{12}x^{12}$, then the value of $a_2+a_4+a_6+\ldots+a_{12}$ is

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