The value of $$-{}^{15}C_1 + 2 \cdot {}^{15}C_2 - 3 \cdot {}^{15}C_3 + \ldots - 15 \cdot {}^{15}C_{15} + {}^{14}C_1 + {}^{14}C_3 + {}^{14}C_5 + \ldots + {}^{14}C_{11}$$ is equal to

We need to find the value of $$-\binom{15}{1} + 2\binom{15}{2} - 3\binom{15}{3} + \ldots - 15\binom{15}{15} + \binom{14}{1} + \binom{14}{3} + \binom{14}{5} + \ldots + \binom{14}{11}$$.

Consider the first part: $$S_1 = \sum_{r=1}^{15} (-1)^r \cdot r \cdot \binom{15}{r}$$.

We know that $$r \cdot \binom{15}{r} = 15 \cdot \binom{14}{r-1}$$. So $$S_1 = 15 \sum_{r=1}^{15} (-1)^r \binom{14}{r-1} = 15 \sum_{j=0}^{14} (-1)^{j+1} \binom{14}{j} = -15 \sum_{j=0}^{14} (-1)^j \binom{14}{j}$$.

Now $$\sum_{j=0}^{14} (-1)^j \binom{14}{j} = (1-1)^{14} = 0$$. So $$S_1 = 0$$.

For the second part: $$S_2 = \binom{14}{1} + \binom{14}{3} + \binom{14}{5} + \ldots + \binom{14}{11}$$.

The sum of all odd-indexed binomial coefficients of order 14 is $$\binom{14}{1} + \binom{14}{3} + \binom{14}{5} + \binom{14}{7} + \binom{14}{9} + \binom{14}{11} + \binom{14}{13} = 2^{13}$$.

Our sum $$S_2$$ is missing $$\binom{14}{13} = 14$$. So $$S_2 = 2^{13} - 14$$.

The total value is $$S_1 + S_2 = 0 + 2^{13} - 14 = 2^{13} - 14$$.

Hence, the correct answer is Option D.