Revision as of 23:17, 26 November 2023 by Admin (Created page with "'''Solution: B''' <math>4 \frac{1-v^{2 n}}{i}=5 \frac{1-v^n}{i}</math> so <math>4 \frac{\left(1-v^n\right)\left(1+v^n\right)}{i}=5 \frac{1-v^n}{i}</math>. If <math>v^n=1</math> then <math>v=1</math> so <math>i=1</math> which cannot give <math>4 a_{\overline{2 n} \mid i}=5 a_{\bar{n} \mid i}</math>. Thus <math>4\left(1+v^n\right)=5</math> so <math>v^n=.25</math> so <math>(1+i)^n=4</math> so <math>(1+i)^{n / 2}=2</math> at which point money is doubled. Thus <math>n / 2</m...")
Exercise
Nov 26'23
Answer
Solution: B
[math]4 \frac{1-v^{2 n}}{i}=5 \frac{1-v^n}{i}[/math] so [math]4 \frac{\left(1-v^n\right)\left(1+v^n\right)}{i}=5 \frac{1-v^n}{i}[/math]. If [math]v^n=1[/math] then [math]v=1[/math] so [math]i=1[/math] which cannot give [math]4 a_{\overline{2 n} \mid i}=5 a_{\bar{n} \mid i}[/math]. Thus [math]4\left(1+v^n\right)=5[/math] so [math]v^n=.25[/math] so [math](1+i)^n=4[/math] so [math](1+i)^{n / 2}=2[/math] at which point money is doubled. Thus [math]n / 2[/math] is the answer.
References
Hlynka, Myron. "University of Windsor Old Tests 62-392 Theory of Interest". web2.uwindsor.ca. Retrieved November 23, 2023.