Revision as of 12:05, 20 November 2023 by Admin (Created page with "'''Solution: D''' The accumulated value of the first year of payments is <math display = "block">2000 s_{\overline{12}|0.005} = 24, 671.12</math>. This amount increases at 2% per year. The effective annual interest rate is 1.005<sup>12</sup> -1 = 0.061678. The present value is then <math display = "block"> \begin{align*} P=24,671.12\sum_{k=1}^{2s}1.02^{k-1}(1.061678)^{-k}=24,671.12\frac{1}{1.02}\sum_{k=1}^{2s}\Biggl(\frac{1.02}{1.061678} \Biggr)^{k} \\ =24,187.37{\frac...")
Exercise
Nov 20'23
Answer
Solution: D
The accumulated value of the first year of payments is
[[math]]2000 s_{\overline{12}|0.005} = 24, 671.12[[/math]]
. This amount increases at 2% per year. The effective annual interest rate is 1.00512 -1 = 0.061678. The present value is then
[[math]]
\begin{align*}
P=24,671.12\sum_{k=1}^{2s}1.02^{k-1}(1.061678)^{-k}=24,671.12\frac{1}{1.02}\sum_{k=1}^{2s}\Biggl(\frac{1.02}{1.061678}
\Biggr)^{k} \\
=24,187.37{\frac{0.960743-0.960743^{26}}{1-0.960743}}=374,444.
\end{align*}
[[/math]]
This is 56 less than the lump sum amount.