Search any question & find its solution
Question:
Answered & Verified by Expert
If $n$ is a positive integer and $\omega \neq 1$ is a cube root of unity, the number of possible values of
$$
\sum_{\mathrm{e} \mathrm{k}=0}^{\mathrm{n}}\left(\begin{array}{l}
\mathrm{n} \\
\mathrm{k}
\end{array}\right) \omega^{\mathrm{k}}
$$
is
Options:
$$
\sum_{\mathrm{e} \mathrm{k}=0}^{\mathrm{n}}\left(\begin{array}{l}
\mathrm{n} \\
\mathrm{k}
\end{array}\right) \omega^{\mathrm{k}}
$$
is
Solution:
1239 Upvotes
Verified Answer
The correct answer is:
4
$\left|e^{\sum_{R=0}^{n}\left(\frac{n}{k}\right) w^{k}}\right|=\left|e^{(1+w)^{n}}\right|=\left|e^{\left(-w^{2}\right)^{n}}\right|$
$\therefore$ when $\mathrm{n}$ is multiplying of 3, then $\mathrm{S}=\mathrm{e}, 1 / \mathrm{e}$
when $\mathrm{n}$ is not multiple of three,then $\mathrm{s}=\mathrm{e}^{1 / 2} \mathrm{e}^{-1 / 2}$
$\therefore$ when $\mathrm{n}$ is multiplying of 3, then $\mathrm{S}=\mathrm{e}, 1 / \mathrm{e}$
when $\mathrm{n}$ is not multiple of three,then $\mathrm{s}=\mathrm{e}^{1 / 2} \mathrm{e}^{-1 / 2}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.