Exercise


Nov 26'23

Answer

Solution: D

[[math]] w=a_{\overline{n} \mid} ; x=a_{\overline{n} \mid} v^n ; y=a_{\overline{n} \mid} v^{2 n} \cdot z=v^{3 n+1}+v^{3 n+2}+\cdots=\frac{v^{3 n+1}}{1-v} [[/math]]


Now

[[math]]w-2 x=a_{\overline{n} \mid}\left(1-2 v^n\right)=z-y=\frac{v^{3 n+1}}{1-v}-a_{\overline{n} \mid} v^{2 n}[[/math]]

. Solve for [math]a_{\overline{n} \mid}[/math] to get

[[math]]a_{\overline{n} \mid}\left(1-2 v^n+v^{2 n}\right)=\frac{v^{3 n+1}}{1-v}[[/math]]

so

[[math]]\frac{\left(1-v^n\right)^3}{i}=\frac{v^{3 n+1}}{i /(1+i)}=\frac{v^{3 n}}{i}[[/math]]

. Thus [math]\left(1-v^n\right)^3=v^{3 n}[/math] so [math]1-v^n=v^n[/math] so [math]v^n=.5[/math]. Finally

[[math]]w / z=\frac{a_{\overline{n} \mid}}{v^{3 n+1 /(1-v)}}=\frac{1-v^n}{v^{3 n}}=\frac{1-.5}{.5^3}=4[[/math]]

.

References

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

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