The sum of the last eight coefficients in the expansion of $(1 + x)^{15}$ is

Let $\alpha = \sum_{k=0}^n \left( \frac{({ }^n C_k)^2}{k+1} \right)$ and $\beta = \sum_{k=0}^{n-1} \left( \frac{{ }^n C_k \cdot { }^n C_{k+1}}{k+2} \right)$. If $5 \alpha = 6 \beta$,then $n$ equals:

If ${}^n C_0, {}^n C_1, {}^n C_2, \ldots, {}^n C_n$ are the binomial coefficients in the expansion of $(1+x)^n$,then for $n=10$,the value of $\sum_{r=1}^{10} {}^n C_r \cdot r(r-4)$ is:

Suppose $\sum_{r=0}^{2023} r \cdot ^{2023}C_r = 2023 \times \alpha \times 2^{2022}$. Then the value of $\alpha$ is $............$

$\binom{50}{4} + \sum_{i=1}^{6} \binom{56-i}{3} = \dots$

If $C_r = ^{100}C_r$,then the value of $1 \cdot C_0^2 - 2 \cdot C_1^2 + 3 \cdot C_2^2 - 4 \cdot C_3^2 + \dots + 101 \cdot C_{100}^2$ is equal to: