The value of ${ }^{10} C_{1}+{ }^{10} C_{2}+{ }^{10} C_{3}+\ldots+{ }^{10} C_{9}$ is

In the expansion of $(1 + x)^n$,the sum of the coefficients of odd powers of $x$ is:

The sum of the series $1 + \frac{1}{2} {}^{n}C_{1} + \frac{1}{3} {}^{n}C_{2} + \dots + \frac{1}{n+1} {}^{n}C_{n}$ is equal to

Let $X = 1({ }^{10} C _1)^2 + 2({ }^{10} C _2)^2 + 3({ }^{10} C _3)^2 + \ldots + 10({ }^{10} C _{10})^2$,where ${ }^{10} C _{ r }$ for $r \in \{1, 2, \ldots, 10\}$ denotes binomial coefficients. Then,the value of $\frac{1}{1430} X$ is:

Let $m, n \in \mathbb{N}$ and $\operatorname{gcd}(2, n)=1$. If $30\binom{30}{0} + 29\binom{30}{1} + \ldots + 2\binom{30}{28} + 1\binom{30}{29} = n \cdot 2^m$,then $n + m$ is equal to (Here $\binom{n}{k} = {^nC_k}$)

If $\binom{40}{0} + \binom{41}{1} + \binom{42}{2} + \dots + \binom{60}{20} = \frac{m}{n} \binom{60}{20}$,where $m$ and $n$ are coprime,then $m+n$ is equal to