Summation Involving Binomial Coefficients and Powers of Two

To find the value of $\alpha$, let us first evaluate the expression inside the parentheses:

$S = \frac{2^3}{3} \binom{12}{2} + \frac{2^5}{5} \binom{12}{4} + \frac{2^7}{7} \binom{12}{6} + \dots + \frac{2^{13}}{13} \binom{12}{12}$

This series can be written in summation notation as: $S = \sum_{k=1}^{6} \frac{2^{2k+1}}{2k+1} \binom{12}{2k}$

Using the binomial identity: $\frac{1}{r+1} \binom{n}{r} = \frac{1}{n+1} \binom{n+1}{r+1}$

For $n = 12$ and $r = 2k$, we have: $\frac{1}{2k+1} \binom{12}{2k} = \frac{1}{13} \binom{13}{2k+1}$

Substituting this back into the expression for $S$: $S = \sum_{k=1}^{6} \frac{1}{13} \binom{13}{2k+1} 2^{2k+1}$ $13S = \sum_{k=1}^{6} \binom{13}{2k+1} 2^{2k+1}$

Now, consider the binomial expansions of $(1+x)^{13}$ and $(1-x)^{13}$: $(1+x)^{13} - (1-x)^{13} = 2 \left[ \binom{13}{1} x + \binom{13}{3} x^3 + \binom{13}{5} x^5 + \dots + \binom{13}{13} x^{13} \right]$

Substitute $x = 2$ into the expansion: $(1+2)^{13} - (1-2)^{13} = 2 \left[ \binom{13}{1} (2) + \sum_{k=1}^{6} \binom{13}{2k+1} 2^{2k+1} \right]$ $3^{13} - (-1)^{13} = 2 \left[ 13 \times 2 + 13S \right]$ $3^{13} + 1 = 2 (26 + 13S)$ $3^{13} + 1 = 52 + 26S$

Rearranging the terms to isolate $26S$: $26S = 3^{13} + 1 - 52$ $26S = 3^{13} - 51$

Given that $26S = 3^{13} - \alpha$, we compare the two equations: $3^{13} - \alpha = 3^{13} - 51 \implies \alpha = 51$

Thus, $\alpha$ is equal to $51$.