Let $$i = \sqrt{-1}$$. If $$\frac{(-1 + i\sqrt{3})^{21}}{(1 - i)^{24}} + \frac{(1 + i\sqrt{3})^{21}}{(1 + i)^{24}} = k$$, and $$n = [|k|]$$ be the greatest integral part of $$|k|$$. Then $$\sum_{j=0}^{n+5} (j + 5)^2 - \sum_{j=0}^{n+5} (j + 5)$$ is equal to ______.

We first convert each complex number to polar form. We have $$-1 + i\sqrt{3} = 2e^{i \cdot 2\pi/3}$$, so $$(-1 + i\sqrt{3})^{21} = 2^{21} e^{i \cdot 42\pi/3} = 2^{21} e^{i \cdot 14\pi} = 2^{21}$$.

Also, $$1 - i = \sqrt{2}\, e^{-i\pi/4}$$, so $$(1 - i)^{24} = (\sqrt{2})^{24} e^{-i \cdot 6\pi} = 2^{12} \cdot 1 = 2^{12}$$.

Therefore, the first fraction is $$\frac{2^{21}}{2^{12}} = 2^9 = 512$$.

Similarly, $$1 + i\sqrt{3} = 2e^{i\pi/3}$$, so $$(1 + i\sqrt{3})^{21} = 2^{21} e^{i \cdot 7\pi} = 2^{21} e^{i\pi} = -2^{21}$$.

And $$1 + i = \sqrt{2}\, e^{i\pi/4}$$, so $$(1 + i)^{24} = 2^{12} e^{i \cdot 6\pi} = 2^{12}$$.

Therefore, the second fraction is $$\frac{-2^{21}}{2^{12}} = -2^9 = -512$$.

Adding the two fractions: $$k = 512 + (-512) = 0$$, so $$|k| = 0$$ and $$n = [|k|] = 0$$.

Now we compute $$\sum_{j=0}^{n+5}(j+5)^2 - \sum_{j=0}^{n+5}(j+5) = \sum_{j=0}^{5}(j+5)^2 - \sum_{j=0}^{5}(j+5)$$. Substituting $$m = j + 5$$, this becomes $$\sum_{m=5}^{10} m^2 - \sum_{m=5}^{10} m = (25 + 36 + 49 + 64 + 81 + 100) - (5 + 6 + 7 + 8 + 9 + 10) = 355 - 45 = 310$$.

The correct answer is $$310$$.