Let i = - 1 . If - 1 + i 3 21 1 - i 24 + 1 + i 3 21 1 + i 24 = k , and n = k be the greatest integral part…

$k=\frac{1}{2^9}\left[\frac{{\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)}^{21}}{{\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i\right)}^{24}}+\frac{{\left(\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)}^{21}}{{\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\right)}^{24}}\right]$

$k=\frac{1}{512}\left[\frac{{\left(e^{i\frac{2\pi }{3}}\right)}^{21}}{{\left(e^{-\frac{i\pi }{4}}\right)}^{24}}+\frac{{\left(e^{\frac{i\pi }{3}}\right)}^{21}}{{\left(e^{\frac{i\pi }{4}}\right)}^{24}}\right.]$

$k=\frac{1}{512}\left[e^{i\left(14\pi +6\pi \right)}+e^{i\left(7\pi -6\pi \right)}\right]$

$k=\frac{1}{512}\left[e^{20\pi i}+e^{\pi i}\right]$

$k=\frac{1}{512}[1+\left(-1\right)]=0$

$n=\left[\left|k\right|\right]=0$

$\sum _{j=0}^5{\left(j+5\right)}^2-\sum _{j=0}^5\left(j+5\right)$

$=\sum _{j=0}^5\left(j^2+25+10j-j-5\right)$

$=\sum _{j=0}^5\left(j^2+9j+20\right)$

$=\sum _{j=0}^5{j}^2+9\sum _{j=0}^{5}j+20\sum _{j=0}^{5}1$

$=\frac{5\times 6\times 11}{6}+9\left(\frac{5\times 6}{2}\right)+20\times 6$

$=55+135+120$

$=310$