Revision as of 09:54, 18 November 2023 by Admin (Created page with "'''Solution: B''' For the first perpetuity, <math display = "block"> \begin{align*} 32 &=10\left( v^{3}+ v^{6}+\cdot\cdot\cdot\right)=10 v^{3}/\left(1- v^{3}\right)\\ 32-32 v^{3} &=10 v^{3} \\ v^{3} &=32/42. \end{align*} </math> For the second perpetuity <math display = "block"> X= v^{13}+ v^{23}+\cdots= v^{1/3}\,/\,(1- v^{1/3})=(32\,/\,42)^{1/9}\,/\,[1-(32\,/\,42)^{1/9}\,]=32.599. </math> {{soacopyright | 2023 }}")
Exercise
Nov 18'23
Answer
Solution: B
For the first perpetuity,
[[math]]
\begin{align*}
32 &=10\left( v^{3}+ v^{6}+\cdot\cdot\cdot\right)=10 v^{3}/\left(1- v^{3}\right)\\
32-32 v^{3} &=10 v^{3} \\
v^{3} &=32/42.
\end{align*}
[[/math]]
For the second perpetuity
[[math]]
X= v^{13}+ v^{23}+\cdots= v^{1/3}\,/\,(1- v^{1/3})=(32\,/\,42)^{1/9}\,/\,[1-(32\,/\,42)^{1/9}\,]=32.599.
[[/math]]
Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.