If $a$ and $d$ are two complex numbers,then the sum to $(n + 1)$ terms of the following series $a{C_0} - (a + d){C_1} + (a + 2d){C_2} - \dots$ is

If $C_0, C_1, C_2, \ldots, C_n$ are the binomial coefficients in the expansion of $(1+x)^n$,then the value of $\sum_{r=0}^{n} r^3 \cdot C_r$ when $n=5$ is

For $r=0, 1, \ldots, 10$,let $A_{r}, B_{r}$ and $C_{r}$ denote,respectively,the coefficient of $x^{r}$ in the expansions of $(1+x)^{10}$,$(1+x)^{20}$ and $(1+x)^{30}$. Then $\sum_{r=1}^{10} A_r(B_{10} B_r - C_{10} A_r)$ 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 sum to $(n + 1)$ terms of the following series $\frac{C_0}{2} - \frac{C_1}{3} + \frac{C_2}{4} - \frac{C_3}{5} + \dots$ is

If $(1 + x)^n = \sum\limits_{r = 0}^n {{C_r}{x^r}} $,then $\left( {1 + \frac{{{C_1}}}{{{C_0}}}} \right)\left( {1 + \frac{{{C_2}}}{{{C_1}}}} \right)....\left( {1 + \frac{{{C_n}}}{{{C_{n - 1}}}}} \right) = $