Question

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.

Answer 1

$K=\frac{1}{2^{9}}\left[\frac{\left(-\frac{1}{2}+\frac{ i \sqrt{3}}{2}\right)^{21}}{\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} i \right)^{24}}+\frac{\left(\frac{1}{2}+\frac{ i \sqrt{3}}{2}\right)^{21}}{\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} i \right)^{24}}\right]$
$K =\frac{1}{512}\left[\frac{\left( e ^{ i \frac{2 \pi}{3}}\right)^{21}}{\left( e ^{-\frac{ i \pi}{4}}\right)^{24}}+\frac{\left( e ^{\frac{ i \pi}{3}}\right)^{21}}{\left( e ^{\frac{ i \pi}{4}}\right)^{24}}\right]$
$K =\frac{1}{512}\left[ e ^{ i (14 \pi+6 \pi)}+ e ^{ i (7 \pi-6 \pi)}\right]$
$K =\frac{1}{512}\left[ e ^{20 \pi i }+ e ^{\pi i }\right]$
$K =\frac{1}{512}[1+(-1)]=0$
$n =[| k |]=0$
$\sum_\limits{ j =0}^{5}( j +5)^{2}-\sum_\limits{ j =0}^{5}( j +5)$
$\sum_\limits{ j =0}^{5}\left( j ^{2}+25+10 j - j -5\right)$
$\sum_\limits{ j =0}^{5}\left( j ^{2}+9 j +20\right)$
$\sum_\limits{j=0}^{5} j ^{2}+9 \sum_\limits{j=0}^{5} j +20 \sum_\limits{j=0}^{5} 1$
$\frac{5 \times 6 \times 11}{6}+9\left(\frac{5 \times 6}{2}\right)+20 \times 6$
$=55+135+120$
$=310$