Geometric Vectors Homework: Get Corrected Answers

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The discussion centers on a request for corrections on geometric vector homework answers. The original poster has shared images of their work and is seeking feedback on accuracy. Participants inquire about the source of the homework, revealing it comes from a handout rather than a textbook. There is some confusion regarding the nature of the questions being asked, with others noting that the answers are already provided. The thread highlights a collaborative effort to ensure understanding of the homework material.
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Homework Statement


See images. . .
Can you please correct my answers if I get some of them wrong?

Homework Equations


There aren't really any relevant equations to solve the problems

The Attempt at a Solution


See images. . .
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What book are you using? I like these question.
 
rocophysics said:
What book are you using? I like these question.

No textbook, it's a handout sheet from my teacher.
 
What question are you asking? You have all the answers to these problems.
 
HallsofIvy said:
What question are you asking? You have all the answers to these problems.

I'm asking whether or not I got these questions right. . .
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

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