Suppose $\sum \limits_{ r =0}^{2023} r ^{20023} C _{ r }=2023 \times \alpha \times 2^{2022}$. Then the value of $\alpha$ is $............$

If for positive integers $r> 1, n > 2$, the coefficients of the $(3r)^{th}$ and $(r + 2)^{th}$ powers of $x$ in the expansion of $( 1 + x)^{2n}$ are equal, then $n$ is equal to 

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

In the expansion of ${(1 + x)^5}$, the sum of the coefficient of the terms is

Let ${\left( {1 + x + {x^2}} \right)^{20}}\left( {2x + 1} \right) = {a_0} + {a_1}{x^1} + {a_2}{x^2} + ... + {a_{41}}{x^{41}}$ , then $\frac{{{a_0}}}{1} + \frac{{{a_1}}}{2} + .... + \frac{{{a_{41}}}}{{42}}$ is equal to 

Let $K$ be the sum of the coefficients of the odd powers of $x$ in the expansion of $(1+ x )^{99}$. Let a be the middle term in the expansion of $\left(2+\frac{1}{\sqrt{2}}\right)^{200}$. If $\frac{{ }^{200} C _{99} K }{ a }=\frac{2^{\ell} m }{ n }$, where $m$ and $n$ are odd numbers, then the ordered pair $(l, n )$ is equal to :