$C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \dots + C_{n-r} C_n =$

$2 \cdot {}^{20}C_0 + 5 \cdot {}^{20}C_1 + 8 \cdot {}^{20}C_2 + 11 \cdot {}^{20}C_3 + \dots + 62 \cdot {}^{20}C_{20}$ is equal to

In the expansion of $(1 + x)^{50}$,the sum of the coefficients of odd powers of $x$ is

$^{15}C_3 + ^{15}C_5 + \ldots + ^{15}C_{15} = ?$

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 $S_2 = 10 \times 2^8$

If $(1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + .... + a_{2n}x^{2n}$,then $a_0 + a_2 + a_4 + .... + a_{2n} = $