If $$26\left(\frac{2^3}{3} {^{12} C_{2}} + \frac{2^5}{5} {^{12} C_{4}} + \frac{2^7}{7} {^{12} C_{6}} + \cdots + \frac{2^{13}}{13} {^{12} C_{12}}\right) = 3^{13} - \alpha$$, then $$\alpha$$ is equal to :

A

45

B

48

C

51

D

54