Revision as of 14:26, 19 November 2023 by Admin (Created page with "'''Solution: D''' <math display = "block"> \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> {{soacopyright | 2023 }}")
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.