JEE Main 2021 (Online) 24th February Evening ShiftQuestion 1
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 $\_\_\_\_\_\_$ .
Complex Numbers
Answer$310$
Solution
$-1+i\sqrt3=2e^{2\pi i/3}$, $(1-i)^{24}=(-2i)^{12}=2^{12}$. First term: $\frac{2^{21}e^{14\pi i}}{2^{12}}=2^9$. $1+i\sqrt3=2e^{i\pi/3}$, $(1+i)^{24}=(2i)^{12}=2^{12}$. Second term: $\frac{2^{21}e^{7\pi i}}{2^{12}}=-2^9$. So $k=0$, $n=0$. The required sum is $\sum_{j=0}^{5}[(j+5)^2-(j+5)]=\sum (j+5)(j+4)=20+30+42+56+72+90=310$.