Exercise


ABy Admin
Nov 26'23

Answer

Solution: E

[[math]] X=100\left[(1+i)^{40}+(1+i)^{36}+\cdots+(1+i)^4\right]=\frac{\left.100 *(1+i)^4 *\left(1-(1+i)^{40}\right)\right)}{1-(1+i)^4} [[/math]]

[[math]] Y=A(20)=100\left[(1+i)^{20}+(1+i)^{16}+\cdots+(1+i)^4\right]=\frac{100 *(1+i)^4 *\left(1-(1+i)^{20}\right)}{1-(1+i)^4} . [[/math]]


Using [math]X=5 Y[/math] and using a difference of squares on the left side gives [math]1+(1+i)^{20}=5[/math] so [math](1+i)^{20}=4[/math] so [math](1+i)^4=4^2=1.319508[/math].

Hence

[[math]]X=\frac{\left.100 *(1+i)^4 *\left(1-(1+i)^{40}\right)\right)}{1-(1+i)^4}=\frac{100 * 1.3195 *\left(1-4^2\right)}{1-1.3195}=6194.84[[/math]]

References

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

00