A perpetuity pays 1 at the end of every year plus an additional 1 at the end of every second year. The present value of the perpetuity is K for i > 0.

Determine K.

• [[math]]\frac{i+3}{i(i+2)}[[/math]]
• [[math]]\frac{i+2}{i(i+1)}[[/math]]
• [[math]]\frac{i+1}{i^2}[[/math]]
• [[math]]\frac{3}{2 i}[[/math]]
• [[math]]\frac{i+1}{i(i+2)}[[/math]]

References

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

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.