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 sum of the series $\binom{20}{0} - \binom{20}{1} + \binom{20}{2} - \binom{20}{3} + \dots + \binom{20}{10}$ is:

For $2 \le r \le n$,$\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}$ is equal to

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 mean of the values $0, 1, 2, \dots, n$ having corresponding weights $^nC_0, ^nC_1, ^nC_2, \dots, ^nC_n$ respectively is

If $n$ is a positive integer, the value of $(2n+1) ^nC_0 + (2n-1) ^nC_1 + (2n-3) ^nC_2 + \ldots + 1 \cdot ^nC_n$ is