Revision as of 08:42, 18 November 2023 by Admin (Created page with "'''Solution: C''' <math display="block"> \begin{align*} 77.1 &=\nu\Big(I a\Big)_{{\overline{n}|i}}^{}+\frac{n\nu^{n+1}}{i} \\ &=\nu\left[\frac{\ddot{a}_{\overline{{{n}}}}-n\nu^{n}}{i}\right]+\frac{n\nu^{n+1}}{i} \\ &=\frac{a_{\overline{n}|i}}{i}-\frac{n\nu^{n+1}}{i}+\frac{n\nu^{n+1}}{i} \\ &=\frac{a_{\overline{n}|i}}{i}=\frac{1-\nu^{n}}{i^{2}}=\frac{1-\nu^{n}}{0.01\ 1\vert025} \\ 0.85003 &=1-\nu^{n} \\ 1.105^{-n} &=0.14997 \\ n &= -\frac{\ln(0.14997)}{\ln(1.105)} = 19 \...")
Exercise
ABy Admin
Nov 18'23
Answer
Solution: C
[[math]]
\begin{align*}
77.1 &=\nu\Big(I a\Big)_{{\overline{n}|i}}^{}+\frac{n\nu^{n+1}}{i} \\
&=\nu\left[\frac{\ddot{a}_{\overline{{{n}}}}-n\nu^{n}}{i}\right]+\frac{n\nu^{n+1}}{i} \\
&=\frac{a_{\overline{n}|i}}{i}-\frac{n\nu^{n+1}}{i}+\frac{n\nu^{n+1}}{i} \\
&=\frac{a_{\overline{n}|i}}{i}=\frac{1-\nu^{n}}{i^{2}}=\frac{1-\nu^{n}}{0.01\ 1\vert025} \\
0.85003 &=1-\nu^{n} \\
1.105^{-n} &=0.14997 \\
n &= -\frac{\ln(0.14997)}{\ln(1.105)} = 19
\end{align*}
[[/math]]
To obtain the present value without remembering the formula for an increasing annuity, consider the payments as a perpetuity of 1 starting at time 2, a perpetuity of 1 starting at time 3, up to a perpetuity of 1 starting at time n + 1. The present value one period before the start of each perpetuity is 1/i. The total present value is
[[math]](1/i)(v+v+v^2+\cdots + v^n) = (1/i)a_{\overline{n}|i}.[[/math]]