Find the arithmetic mean of $^nC_0, ^nC_1, ^nC_2, \dots, ^nC_n$.

If $(1 + x)^{15} = C_0 + C_1x + C_2x^2 + ...... + C_{15}x^{15},$ then $C_2 + 2C_3 + 3C_4 + .... + 14C_{15} = $

If the coefficients of $x^4, x^5$ and $x^6$ in the expansion of $(1+x)^n$ are in arithmetic progression,then the maximum value of $n$ is:

$\sum \limits_{k=0}^{6} {}^{51-k}C_{3}$ is equal to

$\frac{{^nC_0}}{1} + \frac{{^nC_2}}{3} + \frac{{^nC_4}}{5} + \frac{{^nC_6}}{7} + \dots = $

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 $S_2 = 10 \times 2^8$