Revision as of 22:53, 17 November 2023 by Admin (Created page with "'''Solution: A''' <math display = "block"> \begin{align*} 1{\mathrm{00}}{\bigg(}1+{\frac{i}{2}}{\bigg)}^{22}\left[\left(1+{\frac{i}{2}}\right)^{2}-1\right]=2{\bigg(}1{\mathrm00}{\bigg)}{\bigg(}1+{\frac{i}{2}}{\bigg)}^{8}\left[{\bigg(}1+{\frac{i}{2}}{\bigg)}^{2}-1\right] \\ 2=(1+i)^{14}\\ i = (2^{1/14}-1)2 = 0.1015 = 10.15\%. \end{align*} </math> {{soacopyright | 2023 }}")
Exercise
ABy Admin
Nov 17'23
Answer
Solution: A
[[math]]
\begin{align*}
1{\mathrm{00}}{\bigg(}1+{\frac{i}{2}}{\bigg)}^{22}\left[\left(1+{\frac{i}{2}}\right)^{2}-1\right]=2{\bigg(}1{\mathrm00}{\bigg)}{\bigg(}1+{\frac{i}{2}}{\bigg)}^{8}\left[{\bigg(}1+{\frac{i}{2}}{\bigg)}^{2}-1\right] \\
2=(1+i)^{14}\\
i = (2^{1/14}-1)2 = 0.1015 = 10.15\%.
\end{align*}
[[/math]]
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