n: número de periódos de 3 meses.

a0: valor acumulado de la inversión.

Rédito: 8%

Periódo trimestral: 4 al año

Periódo total:

\[4 \times 15 = 60 \hspace{1mm} \texttt{Trimestres} \] \[\texttt{Redito} = \frac{8\%}{4 \hspace{1mm} \texttt{Periodos}} = 2\% \] \[a_{0} = A = ?\] \[a_{60} = 7218.27\] \[a_{n+1} = a_{n} + 0.02a_{n}\] \[a_{n+1} = 1.02a_{n} \geq 0\] \[a_{0} = A\] \[a_{n} = A\left(1.02 \right)^{n}, n \geq 0\] \[a_{60} = 7218.27\] \[7218.27 = A \left(1.02 \right)^{60}\] \[A = \frac{7218.27}{\left( 1.02 \right)^{60}}\] \[A = 2200.00\] \[\texttt{Inversion inicial:} \hspace{1mm} Q2200.00\]

\[\texttt{redito} = 6\%\] \[\texttt{Periodos} = 4\] \[\texttt{r} = \frac{6\%}{4} = 1.5\%\] \[a_{0} = \texttt{inversion} = \$100\] \[\texttt{n} = \texttt{numero de periodos}\] \[\texttt{ecuacion 1}\] \[a_{n+1} = a_{n} + 0.015a_{n}\] \[a_{n+1} = 1.015a_{n}\] \[a_{n} = 100\left( 1.015 \right)^{n}, n \geq 0\] \[a_{n} = 200\] \[A = 100\] \[a_{n} = A\left( 1.015 \right)^{n}\] \[\texttt{Encontrando n:}\] \[a_{n} = 100\left( 1.015 \right)^{n}\] \[200 = 100\left( 1.015 \right)^{n}\] \[2 = \left( 1.015 \right)^{n}\] \[ln{2} = ln{\left( 1.015 \right)^{n}}\] \[n = \frac{ln 2}{ln{1.015}}\] \[n = 46.555\] \[\texttt{n = tiempo}\] \[t = \lceil 46.55 \rceil\] \[t = 47 \times 3\] \[t = 141\] \[t = 141 \texttt{meses}\]