Revision as of 17:45, 26 November 2023 by Admin (Created page with "'''Solution: A''' <math display="block"> \begin{aligned} & 200 v^n+100 v^{2 n}=200 \text { so } 0=x^2+2 x-2 \text { so } v^n=x=\frac{-2 \pm \sqrt{4+8}}{2}=-1+\sqrt{3} \text {. But } v=1 /(1+i) \text { so } \\ & i=(1 / v)-1=\left(\frac{1}{\sqrt{3}-1}\right)^{1 / n}-1=\left(\frac{\sqrt{3}+1}{3-1}\right)^{1 / n}-1 \end{aligned} </math> '''References''' {{cite web |url=https://web2.uwindsor.ca/math/hlynka/392oldtests.html |last=Hlynka |first=Myron |website=web2.uwindsor...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Exercise


ABy Admin
Nov 26'23

Answer

Solution: A

[[math]] \begin{aligned} & 200 v^n+100 v^{2 n}=200 \text { so } 0=x^2+2 x-2 \text { so } v^n=x=\frac{-2 \pm \sqrt{4+8}}{2}=-1+\sqrt{3} \text {. But } v=1 /(1+i) \text { so } \\ & i=(1 / v)-1=\left(\frac{1}{\sqrt{3}-1}\right)^{1 / n}-1=\left(\frac{\sqrt{3}+1}{3-1}\right)^{1 / n}-1 \end{aligned} [[/math]]

References

Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.

00