Let i=sqrt{-1}. If frac{(-1+i sqrt{3})^{21}}{ | vedclass

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(B) Given $k = \frac{(-1+i \sqrt{3})^{21}}{(1-i)^{24}}+\frac{(1+i \sqrt{3})^{21}}{(1+i)^{24}}$.We know that $-1+i \sqrt{3} = 2(\cos \frac{2\pi}{3} + i

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$310$

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(B) Given $k = \frac{(-1+i \sqrt{3})^{21}}{(1-i)^{24}}+\frac{(1+i \sqrt{3})^{21}}{(1+i)^{24}}$.We know that $-1+i \sqrt{3} = 2(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}) = 2e^{i \frac{2\pi}{3}}$ and $1+i \sqrt{3} = 2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) = 2e^{i \frac{\pi}{3}}$.Also $1-i = \sqrt{2}(\cos(-\frac{\pi}{4}) + i \sin(-\frac{\pi}{4})) = \sqrt{2}e^{-i \frac{\pi}{4}}$ and $1+i = \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}) = \sqrt{2}e^{i \frac{\pi}{4}}$.Substituting these values:$k = \frac{(2e^{i \frac{2\pi}{3}})^{21}}{(\sqrt{2}e^{-i \frac{\pi}{4}})^{24}} + \frac{(2e^{i \frac{\pi}{3}})^{21}}{(\sqrt{2}e^{i \frac{\pi}{4}})^{24}} = \frac{2^{21} e^{i 14\pi}}{2^{12} e^{-i 6\pi}} + \frac{2^{21} e^{i 7\pi}}{2^{12} e^{i 6\pi}}$$k = 2^9 (e^{i 20\pi} + e^{i \pi}) = 512(1 - 1) = 0$.Thus,$n = [|k|] = 0$.The expression becomes $\sum_{j=0}^{5} (j+5)^2 - \sum_{j=0}^{5} (j+5) = \sum_{j=0}^{5} ((j+5)^2 - (j+5)) = \sum_{j=0}^{5} (j^2 + 10j + 25 - j - 5) = \sum_{j=0}^{5} (j^2 + 9j + 20)$.$= \sum_{j=0}^{5} j^2 + 9 \sum_{j=0}^{5} j + \sum_{j=0}^{5} 20 = \frac{5(6)(11)}{6} + 9 \frac{5(6)}{2} + 20(6) = 55 + 135 + 120 = 310$.

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 ........ .

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