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]]

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

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