The sum $\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \ldots$ up to 8 terms, is :

If for $3 \leq r \leq 30$, $\left({^{30}C_{30-r}}\right) + 3\left({^{30}C_{31-r}}\right) + 3\left({^{30}C_{32-r}}\right) + \left({^{30}C_{33-r}}\right) = {^mC_r}$, then m equals :

Let $p_n$ denote the total number of triangles formed by joining the vertices of an $n$-side regular polygon.
If $p_{n+1} - p_n = 66$, then the sum of all distinct prime divisors of $n$ is :

A man throws a fair coin repeatedly. He gets 10 points for each head he throws and 5 points for each tail he throws. If the probability that he gets exactly 30 points is $\frac{m}{n}$, gcd $(m, n) = 1$, then m + n is equal to :