QJEEM21Mab1d2: JEE Main 2021 Mathematics - Class 11 JEEM21Mab1d2 Solution: Let . If , and be the greatest integral part of

Question

Verified Solution • 2025–26 Session moderate

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 _________.

Accepted Answer

{(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

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