be the greatest integral part of |k|

Question:

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 $\mathrm{n}=[|k|]$ be the greatest integral part of $|\mathrm{k}|$. Then $\sum_{j=0}^{n+5}(j+5)^{2}-\sum_{j=0}^{n+5}(j+5)$ is equal to

Solution:

$\frac{\left(2 e^{i \frac{2 \pi}{3}}\right)^{21}}{\left(\sqrt{2} e^{-i \frac{\pi}{4}}\right)^{21}}+\frac{\left(2 e^{i \frac{\pi}{3}}\right)^{21}}{\left(\sqrt{2} e^{i \frac{\pi}{4}}\right)^{24}}$

$\Rightarrow \frac{2^{21} \cdot e^{14 \pi}}{2^{12} \cdot e^{-i 6 \pi}}+\frac{2^{21}\left(e^{i 7 \pi}\right)}{2^{12}\left(e^{i 6 \pi}\right)}$

$\Rightarrow 2^{9} e^{i(20 \pi)}+2^{9} e^{i \pi}$

$\Rightarrow 2^{9}+2^{9}(-1)=0$

$n=0$

$\sum_{j=0}^{5}(j+5)^{2}-\sum_{j=0}^{5}(j+5)$

$\Rightarrow\left[5^{2}+6^{2}+7^{2}+8^{2}+9^{2}+10^{2}\right]-[5+6+7+8+9+10]$

$\Rightarrow\left[\left(1^{2}+2^{2}+\ldots+10^{2}\right)-\left(1^{2}+2^{2}+3^{2}+4^{2}\right)\right]-[(1+2+3+\ldots+10)-(1+2+3+4)]$

$\Rightarrow(385-30)-[55-10]$

$\Rightarrow 355-45 \Rightarrow 310$ ans.

 

Leave a comment