Show that the series sum_r=0^infinity (1)/(4^r) 2rtheta is a convergent geometric progression.

May 27, 2026Mathematics

ScanSolve AI Answer

May 27, 2026

Answer

\frac{1}{4}(\cos 2\theta + i\sin 2\theta)$.

Here's the solution for question 6b:

Part (i) To show that the series $\sum_{r=0}^{\infty} \frac{1}{4^r} \sin 2r\theta$ is a convergent geometric progression, we consider a related complex geometric series. Let $w = \frac{1}{4}(\cos 2\theta + i\sin 2\theta)$ . By De Moivre's theorem, $w^r = \left(\frac{1}{4}\right)^r (\cos 2r\theta + i\sin 2r\theta) = \frac{1}{4^r} \cos 2r\theta + i \frac{1}{4^r} \sin 2

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Here's the solution for question 6b: Part (i) To show that the series _r=0^ (1)/(4^r) 2r is a convergent geometric progression, we consider a related complex geometric series.

Asked on May 27, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

ScanSolve AI AnswerAnswered on May 27, 2026

Here's the solution for question 6b: Part (i) To show that the series _r=0^ (1)/(4^r) 2r is a convergent geometric progression, we consider a related complex geometric series. Let w = (1)/(4)( 2 + i 2). By De Moivre's theorem, $w^r = ((1)/(4))^r ( 2r + i 2r) = (1)/(4^r) 2r + i (1)/(4^r) 2