Question

If n is a positive integer and $\omega \ne 1$ is a cube root of unity, the number of possible values of $\sum _{k=0}^n\left[\begin{array}{c}n\\ k\end{array}\right]{\omega }^k$ is$?$

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

The correct option is A $2$

$\sum _{k=0}^n{}_k^{n}C{\omega }^k={}_0^{n}C{\omega }^0+{}_1^{n}C{\omega }^1+{}_2^{n}C{\omega }^2...........{}_k^{n}C{\omega }^k\sum _{k=0}^n{}_k^{n}C{\omega }^k={}_0^{n}C+{}_1^{n}C{\omega }^1+{}_2^{n}C{\omega }^2...........{}_k^{n}C{\omega }^k{(1+x)}^n={}_0^{n}C+{}_1^{n}C{x}^1+{}_2^{n}C{x}^2...........{}_k^{n}C{x}^k\therefore \sum _{k=0}^n{}_k^{n}C{\omega }^k={(1+\omega )}^n=(-{\omega }^2{)}^n={(-1)}^n{\omega }^{2n}\left|e^{{\sum }_{k=0}^n{}_k^{n}C{\omega }^k}\right|=\left|e^{{(-1)}^n{\omega }^{2n}}\right|=\left|e^{{(-{\omega }^2)}^n}\right|{\omega }^2=\frac{-1}{2}-i\frac{\sqrt{3}}{2}{\omega }^2=-cos\frac{4\pi }{3}-isin\frac{4\pi }{3}\left|e^{{\sum }_{k=0}^n{}_k^{n}C{\omega }^k}\right|=\left|e^{{-(-cos\frac{4\pi }{3}-isin\frac{4\pi }{3})}^n}\right|=\left|e^{{(cos\frac{4n\pi }{3}+isin\frac{4n\pi }{3})}^{}}\right|$