If $\binom{10}{2} + \binom{10}{3} + \binom{11}{4} + \binom{12}{5} + \binom{13}{6} = \binom{14}{r}$,then $r = \dots$

If $26 \left( \frac{2}{3} \binom{12}{2} + \frac{2}{5} \binom{12}{4} + \frac{2}{7} \binom{12}{6} + \dots + \frac{2}{13} \binom{12}{12} \right) = 3^{13} - \alpha$, then $\alpha$ is equal to:

The value of $\sum\limits_{r = 0}^{n - 1} {\frac{{^n{C_r}}}{{^n{C_r} + {\,^n}{C_{r + 1}}}}} $ equals

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 the coefficients of $a^{r-1}$,$a^{r}$,and $a^{r+1}$ in the expansion of $(1+a)^{n}$ are in arithmetic progression,prove that $n^{2}-n(4r+1)+4r^{2}-2=0$.

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