Revision as of 21:46, 17 November 2023 by Admin (Created page with "'''Solution: E''' From basic principles, the accumulated values after 20 and 40 years are <math display="block"> \begin{align*} 100[(1+i)^{20}+(1+i)^{16}+\cdots+(1+i)^{4}]=100\frac{(1+i)^{4}-(1+i)^{24}}{1-(1+i)^{4}} \\ 100[(1+i)^{40}+(1+i)^{36}+\cdots+(1+i)^{4}]=100{\frac{(1+i)^{4}-(1+i)^{44}}{1-(1+i)^{4}}} \end{align*} </math> The ratio is 5, and thus (setting <math>x=(1+i)^4</math>)) <math display="block"> \begin{array}{l}{{5=\frac{(1+i)^{4}-(1+i)^{44}}{(1+i)^{4}-(...")
Exercise
ABy Admin
Nov 17'23
Answer
Solution: E
From basic principles, the accumulated values after 20 and 40 years are
[[math]]
\begin{align*}
100[(1+i)^{20}+(1+i)^{16}+\cdots+(1+i)^{4}]=100\frac{(1+i)^{4}-(1+i)^{24}}{1-(1+i)^{4}} \\
100[(1+i)^{40}+(1+i)^{36}+\cdots+(1+i)^{4}]=100{\frac{(1+i)^{4}-(1+i)^{44}}{1-(1+i)^{4}}}
\end{align*}
[[/math]]
The ratio is 5, and thus (setting [math]x=(1+i)^4[/math]))
[[math]]
\begin{array}{l}{{5=\frac{(1+i)^{4}-(1+i)^{44}}{(1+i)^{4}-(1+i)^{24}}=\frac{x-x^{11}}{x-x^{6}}}}\\ {{5x-5x^{5}=x-x^{11}}}\\ {{(x^{5}-1)(x^{5}-4)=0}}\end{array}
[[/math]]
Only the second root gives a positive solution. Thus
[[math]]
\begin{align*}x^{5} &=4\\ x &= 1.31951 \\ x &= 1.31951 \\ X &= 100 \frac{1.31951-1.31951^{11}}{1-1.31951} = 6195.\end{align*}
[[/math]]
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