JEE MAIN - Mathematics (2021 - 24th February Evening Shift - No. 16)
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 _________.
Answer
310
Explanation
$${(1 + i)^2} = 1 + {i^2} + 2i = 1 - 1 + 2i = 2i$$
$${(1 - i)^2} = 1 + {i^2} - 2i = 1 - 1 - 2i = - 2i$$
We know,
$$ - {1 \over 2} + {{i\sqrt 3 } \over 2} = \omega $$
$$ \Rightarrow - 1 + i\sqrt 3 = 2\omega $$
and $$ - {1 \over 2} - {{i\sqrt 3 } \over 2} = {\omega ^2}$$
$$ \Rightarrow - 1 - i\sqrt 3 = 2{\omega ^2}$$
$$ \Rightarrow 1 + i\sqrt 3 = - 2{\omega ^2}$$
Now, $$K = {{{{\left( { - 1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 - i} \right)}^{24}}}} + {{{{\left( {1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 + i} \right)}^{24}}}}$$
$$ = {{{{(2\omega )}^{21}}} \over {{{\left( {{{(1 - i)}^2}} \right)}^{12}}}} + {{{{( - 2\omega )}^{21}}} \over {{{\left( {{{(1 + i)}^2}} \right)}^{12}}}}$$
$$ = {{{2^{21}}.{\omega ^{21}}} \over {{{( - 2i)}^{12}}}} + {{{{( - 2)}^{21}}{{({\omega ^2})}^{21}}} \over {{{(2i)}^{12}}}}$$ [as $${\omega ^3} = 1$$, $${i^4} = 1$$]
$$ = {{{2^{21}}} \over {{2^{12}}}} - {{{2^{21}}} \over {{2^{12}}}} = 0$$
$$ \therefore $$ $$n = \left[ {|K|} \right] = \left[ {|0|} \right] = 0$$
Now $$\sum\limits_{j = 0}^5 {{{(j + 5)}^2}} - \sum\limits_{j = 0}^5 {(j + 5)} $$
= $$\sum\limits_{j = 0}^5 {({j^2} + 25 + 10j - j - 5)} $$
= $$\sum\limits_{j = 0}^5 {({j^2} + 9j + 20)} $$
= $$\sum\limits_{j = 0}^5 {{j^2}} + 9\sum\limits_{j = 0}^5 {j + 20\sum\limits_{j = 0}^5 1 } $$
= $${{5 \times 6 \times 11} \over 6} + 9\left( {{{5 \times 6} \over 2}} \right) + 20 \times 6$$
= 55 + 135 + 120
= 310
$${(1 - i)^2} = 1 + {i^2} - 2i = 1 - 1 - 2i = - 2i$$
We know,
$$ - {1 \over 2} + {{i\sqrt 3 } \over 2} = \omega $$
$$ \Rightarrow - 1 + i\sqrt 3 = 2\omega $$
and $$ - {1 \over 2} - {{i\sqrt 3 } \over 2} = {\omega ^2}$$
$$ \Rightarrow - 1 - i\sqrt 3 = 2{\omega ^2}$$
$$ \Rightarrow 1 + i\sqrt 3 = - 2{\omega ^2}$$
Now, $$K = {{{{\left( { - 1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 - i} \right)}^{24}}}} + {{{{\left( {1 + i\sqrt 3 } \right)}^{21}}} \over {{{\left( {1 + i} \right)}^{24}}}}$$
$$ = {{{{(2\omega )}^{21}}} \over {{{\left( {{{(1 - i)}^2}} \right)}^{12}}}} + {{{{( - 2\omega )}^{21}}} \over {{{\left( {{{(1 + i)}^2}} \right)}^{12}}}}$$
$$ = {{{2^{21}}.{\omega ^{21}}} \over {{{( - 2i)}^{12}}}} + {{{{( - 2)}^{21}}{{({\omega ^2})}^{21}}} \over {{{(2i)}^{12}}}}$$ [as $${\omega ^3} = 1$$, $${i^4} = 1$$]
$$ = {{{2^{21}}} \over {{2^{12}}}} - {{{2^{21}}} \over {{2^{12}}}} = 0$$
$$ \therefore $$ $$n = \left[ {|K|} \right] = \left[ {|0|} \right] = 0$$
Now $$\sum\limits_{j = 0}^5 {{{(j + 5)}^2}} - \sum\limits_{j = 0}^5 {(j + 5)} $$
= $$\sum\limits_{j = 0}^5 {({j^2} + 25 + 10j - j - 5)} $$
= $$\sum\limits_{j = 0}^5 {({j^2} + 9j + 20)} $$
= $$\sum\limits_{j = 0}^5 {{j^2}} + 9\sum\limits_{j = 0}^5 {j + 20\sum\limits_{j = 0}^5 1 } $$
= $${{5 \times 6 \times 11} \over 6} + 9\left( {{{5 \times 6} \over 2}} \right) + 20 \times 6$$
= 55 + 135 + 120
= 310
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