The number $111......1 $ ( $ 91$ times) is

The coefficient of $x ^{101}$ in the expression $(5+x)^{500}+x(5+x)^{499}+x^{2}(5+x)^{498}+\ldots . x^{500}$ $x>0$, is

The value of $4 \{^nC_1 + 4 . ^nC_2 + 4^2 . ^nC_3 + ...... + 4^{n - 1}\}$ is :

Let n and k be positive integers such that $n \ge \frac{{k(k + 1)}}{2}$. The number of solutions $({x_1},{x_2},....{x_k})$, ${x_1} \ge 1,{x_2} \ge 2,....{x_k} \ge k,$ all integers, satisfying ${x_1} + {x_2} + .... + {x_k} = n$, is

Let $(1 + x)^m = C_0 + C_1x + C_2x^2 + C_3x^3 + . . . . . +C_mx^m$,  where $C_r ={}^m{C_r}$ and $A = C_1C_3 + C_2C_4+ C_3C_5 + C_4C_6 + . . . . . .. + C_{m-2}C_m$,  then which is false

Let $C _{ r }$ denote the binomial coefficient of $x ^{ r }$ in the expansion of $(1+x)^{10}$. If $\alpha, \beta \in R$. $C _{1}+3 \cdot 2 C _{2}+5 \cdot 3 C _{3}+\ldots$ upto $10$ terms $=\frac{\alpha \times 2^{11}}{2^{\beta}-1}\left( C _{0}+\frac{ C _{1}}{2}+\frac{ C _{2}}{3}+\ldots . .\right.$ upto 10 terms $)$ then the value of $\alpha+\beta$ is equal to