We need to use the formula:
[tex]A(t)=P(1+\frac{r}{n})^{nt}[/tex]
to find the worth of the account in 20 years.
We know that:
[tex]\begin{gathered} P=6500 \\ r=3.6\%=0.036 \\ n=2 \\ t=20 \end{gathered}[/tex]
Thus, we obtain:
[tex]\begin{gathered} A(20)=6500\cdot(1+\frac{0.036}{2})^{2\cdot20} \\ \\ A(20)=6500\cdot(1+0.018)^{40} \\ \\ A(20)=6500\cdot(1.018)^{40} \\ \\ A(20)=13268.58 \end{gathered}[/tex]
Therefore, the account, after 20 years, will be worth $13268.58.
If the interest were compounded weekly, n would be:
[tex]n=\frac{365}{7}\cong52[/tex]
Then, the account would have been worth:
[tex]\begin{gathered} A(20)=6500\cdot(1+\frac{0.036}{52})^{52\cdot20} \\ \\ A(20)=6500\cdot(1+\frac{0.036}{52})^{1040} \\ \\ A(20)\cong13350.49 \end{gathered}[/tex]
Notice that if we do not approximate the number of weeks (n) to 52, and instead use its exact value (365/7), then we obtain:
[tex]A(20)=6500\cdot\mleft(1+\frac{0.036}{\frac{365}{7}}\mright)^{\frac{365}{7}\cdot20}\cong13350.50[/tex]
If the interest were compounded weekly, the account would have been worth $13350.49 (using n = 52).
