Exercise


ABy Admin
Nov 19'23

Answer

Solution: D

[[math]] \begin{aligned} & I_3+I_4=(1+i)^2\left(P_1+P_2\right) \\ & \left(1-v^2\right)+(1-v)=(1+i)^2\left(v^4+v^3\right) \\ & \left(1-v^2+1-v\right)=(1+i)^2\left(v^4+v^3\right) \\ & v^2\left(1-v^2+1-v\right)=\left(v^4+v^3\right) \\ & 2 v^2-v^2\left(v^2+v\right)=v^2\left(v^2+v\right) \\ & 2 v^2=2 v^2\left(v^2+v\right) \\ & 1=v^2+v \\ & v^2+v-1=0 \\ & v=0.61803 \\ & 1+i=1.61803 \\ & i=0.61803\end{aligned} [[/math]]

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

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