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

${(1 + i)^2} = 1 + {i^2} + 2i = 1 - 1 + 2i = 2i$<br><br>${(1 - i)^2} = 1 + {i^2} - 2i = 1 - 1 - 2i = - 2i$<br><br>We know,<br><br>$- {1 \over 2} + {{i\sqrt 3 } \over 2} = \omega$<br><br>$\Rightarrow - 1 + i\sqrt 3 = 2\omega$<br><br>and $- {1 \over 2} - {{i\sqrt 3 } \over 2} = {\omega ^2}$<br><br>$\Rightarrow - 1 - i\sqrt 3 = 2{\omega ^2}$<br><br>$\Rightarrow 1 + i\sqrt 3 = - 2{\omega ^2}$<br><br>Now, $$K = {{{{\left( { - 1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 - i} \right)}^{24}}}} + {{{{\left( {1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 + i} \right)}^{24}}}}$$<br><br>$$ = {{{{(2\omega )}^{21}}} \over {{{\left( {{{(1 - i)}^2}} \right)}^{12}}}} + {{{{( - 2\omega )}^{21}}} \over {{{\left( {{{(1 + i)}^2}} \right)}^{12}}}}$$<br><br>$$ = {{{2^{21}}.{\omega ^{21}}} \over {{{( - 2i)}^{12}}}} + {{{{( - 2)}^{21}}{{({\omega ^2})}^{21}}} \over {{{(2i)}^{12}}}}$$ [as ${\omega ^3} = 1$, ${i^4} = 1$]<br><br>$= {{{2^{21}}} \over {{2^{12}}}} - {{{2^{21}}} \over {{2^{12}}}} = 0$<br><br>$\therefore$ $n = \left[ {|K|} \right] = \left[ {|0|} \right] = 0$<br><br>Now $\sum\limits_{j = 0}^5 {{{(j + 5)}^2}} - \sum\limits_{j = 0}^5 {(j + 5)}$<br><br>= $\sum\limits_{j = 0}^5 {({j^2} + 25 + 10j - j - 5)}$<br><br>= $\sum\limits_{j = 0}^5 {({j^2} + 9j + 20)}$<br><br>= $$\sum\limits_{j = 0}^5 {{j^2}} + 9\sum\limits_{j = 0}^5 {j + 20\sum\limits_{j = 0}^5 1 } $$<br><br>= $${{5 \times 6 \times 11} \over 6} + 9\left( {{{5 \times 6} \over 2}} \right) + 20 \times 6$$<br><br>= 55 + 135 + 120<br><br>= 310