Question
$1+x^{2}=\sqrt{3} x$, then $\sum_{n=1}^{24}\left(x^{n}-\frac{1}{x^{n}}\right)^{2}$ is equal to(A) 48(B) $-48$ $\begin{array}{ll}(\mathrm{C})+48\left(\omega-\omega^{2}\right) & \text { (D) } 1+48\end{array}$
Step 1
This can be rewritten as $x^{2}-\sqrt{3}x+1=0$. Show more…
Show all steps
Your feedback will help us improve your experience
Mahipal Kumawat and 53 other Algebra educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
$1+x^{2}=\sqrt{3} x$, then $\sum_{n=1}^{24}\left(x^{n}-\frac{1}{x^{n}}\right)^{2}$ is equal to (A) 48 (B) $-48$ $\begin{array}{ll}(\mathrm{C})+48\left(\omega-\omega^{2}\right) & \text { (D) } 1+48\end{array}$
$$ \begin{gathered} \frac{1}{\sqrt{2 x+1}}\left\{(1+\sqrt{2 x+1})^{n}-(1-\sqrt{2 x+1})^{n}\right\} \\ =a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{10} x^{10} \end{gathered} $$ then $n$ must be equal to (a) 20,21 (b) 21,22 (c) 22, 23 (d) 23, 24
If $\omega=\cos \left(\frac{2 \pi}{n}\right)+i \sin \left(\frac{2 \pi}{n}\right)$, then numerical value $1+\frac{1}{2-\omega}+\frac{1}{2-\omega^{2}}+\cdots+\frac{1}{2-\omega^{n-1}}$ equals (a) $\frac{n\left(2^{n-1}\right)}{2^{n}+1}$ (b) $\frac{n\left(2^{n-1}\right)}{2^{n}-1}$ (c) 0 (d) 1
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD