Revision as of 17:36, 26 November 2023 by Admin (Created page with "'''Solution: E''' <math display="block"> 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 display="block"> 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> s...")
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.