Revision as of 23:10, 26 November 2023 by Admin (Created page with "'''Solution: A''' <math>K=v+v^2+\cdots+v^2+v^4+v^6+\cdots=\frac{v}{1-v}+\frac{v^2}{1-v^2}=\frac{(1+v) v}{1-v^2}+\frac{v^2}{1-v^2}=\frac{v+2 v^2}{1-v^2}</math> <math>=\left(\right.</math> mult. num and den by <math>\left.(1+i)^2\right)=\frac{1+i+2}{(1+i)^2-1}=\frac{3+i}{i(2+i)}</math> '''References''' {{cite web |url=https://web2.uwindsor.ca/math/hlynka/392oldtests.html |last=Hlynka |first=Myron |website=web2.uwindsor.ca | title = University of Windsor Old Tests 62-392...")
Exercise
Nov 26'23
Answer
Solution: A
[math]K=v+v^2+\cdots+v^2+v^4+v^6+\cdots=\frac{v}{1-v}+\frac{v^2}{1-v^2}=\frac{(1+v) v}{1-v^2}+\frac{v^2}{1-v^2}=\frac{v+2 v^2}{1-v^2}[/math] [math]=\left(\right.[/math] mult. num and den by [math]\left.(1+i)^2\right)=\frac{1+i+2}{(1+i)^2-1}=\frac{3+i}{i(2+i)}[/math]
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.