The sum to $(n + 1)$ terms of the series $\frac{C_0}{2} - \frac{C_1}{3} + \frac{C_2}{4} - \frac{C_3}{5} + \dots$ is

If the sum of the coefficients of even powers of $x$ in the expansion of $(1-x+x^2)^{2n}$ is $3281$,then $n=$

If $3 \times { }^5 C_0 + 8 \times { }^5 C_1 + 13 \times { }^5 C_2 + 18 \times { }^5 C_3 + 23 \times { }^5 C_4 + 28 \times { }^5 C_5 = k \times 2^n$,where $n$ is a power of $2$,find $k$. Specifically,if $3 \times { }^5 C_0 + 8 \times { }^5 C_1 + 13 \times { }^5 C_2 + 18 \times { }^5 C_3 + 23 \times { }^5 C_4 + 28 \times { }^5 C_5 = k \times 2^4$,then $k=$

If $n$ is a positive integer,then $\sum_{r=1}^n r^2 \cdot C_r = (\ldots \ldots \ldots) 2^{n-2}$

Let ${ }^{n} C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of $(1+ x )^{ n }.$ If $\sum_{ k =0}^{10}\left(2^{2}+3 k \right){ }^{10} C _{ k }=\alpha \cdot 3^{10}+\beta \cdot 2^{10},$ where $\alpha, \beta \in R,$ then $\alpha+\beta$ is equal to ....... .

If $\frac{{}^{11}C_1}{2} + \frac{{}^{11}C_2}{3} + \dots + \frac{{}^{11}C_9}{10} = \frac{n}{m}$ with $\gcd(n, m) = 1$,then $n + m$ is equal to