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

Let $(1+x)^{10} = \sum_{r=0}^{10} c_{r} x^{r}$ and $(1+x)^{7} = \sum_{r=0}^{7} d_{r} x^{r}$. If $P = \sum_{r=0}^{5} c_{2r}$ and $Q = \sum_{r=0}^{3} d_{2r+1}$, then $\frac{P}{Q}$ is equal to:

If $n \geq 2$ is a positive integer,then the sum of the series ${ }^{n+1} C_{2}+2\left({ }^{2} C_{2}+{ }^{3} C_{2}+{ }^{4} C_{2}+\ldots+{ }^{n} C_{2}\right)$ is ...... .

If $C_0, C_1, C_2, \ldots, C_{10}$ represent the binomial coefficients in the expansion of $(1+x)^{10}$,then $C_0 C_6+C_1 C_7+C_2 C_8+C_3 C_9+C_4 C_{10}=$

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 $k=$

The sum of the last $30$ coefficients in the expansion of $(1+x)^{59}$, when expanded in ascending powers of $x$, is: