Exercise


Nov 26'23

Answer

Solution: C

[[math]] \begin{aligned} 77.1 & =1 v^2+2 v^3+3 v^4+\cdots+n v^{n+1}+n\left(v^{n+2}+\ldots\right)=v(I a)_{\overline{n} \mid}+n v^{n+1} a_{\overline{\infty} \mid} \\ & =v \frac{\ddot{a}_{\overline{n} \mid}-n v^n}{i}+n v^{n+1} \frac{1}{i} \\ & =\frac{a_{\overline{n} \mid}-n v^{n+1}}{i}+\frac{n v^{n+1}}{i}=\frac{a_{\overline{n} \mid i}}{i} \\ & =\frac{1-v^n}{i^2} \end{aligned} [[/math]]

Thus [math]v^n=1-77.1\left(i^2\right)[/math] so

[[math]]n=\frac{\ln \left(1-77.1 i^2\right)}{\ln (v)}=\frac{-1.8973}{-.099845}=19[[/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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