If $$\left(\frac{1}{{}^{15}C_0} + \frac{1}{{}^{15}C_1}\right) \left(\frac{1}{{}^{15}C_1} + \frac{1}{{}^{15}C_2}\right) \cdots \left(\frac{1}{{}^{15}C_{12}} + \frac{1}{{}^{15}C_{13}}\right) = \frac{\alpha^{13}}{{}^{14}C_0 {}^{14}C_1 \cdots {}^{14}C_{12}}$$ then $$30\alpha$$ is equal to __________

We need to evaluate the product $$\prod_{r=0}^{12}\left(\frac{1}{\binom{15}{r}} + \frac{1}{\binom{15}{r+1}}\right) = \frac{\alpha^{13}}{\binom{14}{0}\binom{14}{1}\cdots\binom{14}{12}}$$.

We begin by simplifying each factor: $$\frac{1}{\binom{15}{r}} + \frac{1}{\binom{15}{r+1}} = \frac{\binom{15}{r+1} + \binom{15}{r}}{\binom{15}{r}\binom{15}{r+1}} = \frac{\binom{16}{r+1}}{\binom{15}{r}\binom{15}{r+1}}$$

Using Pascal's rule $$\binom{15}{r} + \binom{15}{r+1} = \binom{16}{r+1}$$, we note that $$\binom{16}{r+1} = \frac{16}{r+1}\binom{15}{r}$$.

Substituting this identity gives $$\frac{\binom{16}{r+1}}{\binom{15}{r}\binom{15}{r+1}} = \frac{16}{(r+1)\binom{15}{r+1}}$$

Hence the product becomes $$\prod_{r=0}^{12} \frac{16}{(r+1)\binom{15}{r+1}} = \frac{16^{13}}{\prod_{r=0}^{12}(r+1) \cdot \prod_{r=0}^{12}\binom{15}{r+1}} = \frac{16^{13}}{13! \cdot \prod_{s=1}^{13}\binom{15}{s}}$$

We then use $$\binom{15}{s} = \frac{15!}{s!(15-s)!}$$ and $$\binom{14}{s-1} = \frac{14!}{(s-1)!(15-s)!}$$ to observe that $$\binom{15}{s} = \frac{15}{s}\binom{14}{s-1}$$.

It follows that $$\prod_{s=1}^{13}\binom{15}{s} = \prod_{s=1}^{13}\frac{15}{s}\binom{14}{s-1} = \frac{15^{13}}{13!}\prod_{s=1}^{13}\binom{14}{s-1} = \frac{15^{13}}{13!}\prod_{j=0}^{12}\binom{14}{j}$$

Substituting back yields $$\frac{16^{13}}{13! \cdot \frac{15^{13}}{13!}\prod_{j=0}^{12}\binom{14}{j}} = \frac{16^{13}}{15^{13}\prod_{j=0}^{12}\binom{14}{j}} = \frac{(16/15)^{13}}{\prod_{j=0}^{12}\binom{14}{j}}$$

Therefore $$\alpha = \frac{16}{15}$$ and $$30\alpha = 30 \times \frac{16}{15} = 32$$.

The answer is 32.