Question

Use the numbers from the original example: $1,000 invested at a 2% interest rate compounded n times per year. Compare the change in P as n increases. Fill in the table. Use a calculator and write the values to 5 decimal places. Use commas in your answer when necessary.

n
P=P0(1+r/n)^n
1 (once per year)
a0
4 (every 3 months)
a1
12 (every month)
a2
52 (every week)
a3
365 (every day)
a4

Answer 1

So, we are going to use the compound interest formula: $P=P_{0}(1+ \frac{r}{n} )^{nt}$
where 
$P$ is the final amount 
$P_{0}$ is the initial amount 
$r$ is the interest rate in decimal form 
$n$ is the number of times the interest is compounded per year
$t$ is the time in years 
Since in all the cases $t=1$, we can omit $t$.

1. Once per year. In this case: $P_{0}=1000$, $r= \frac{2}{100} =0.02$, and $n=1$. Lets replace those values in our formula:
$P=1000(1+ \frac{0.02}{1} )^{1}$
$P=1020$

2. Every three months. In this case: $P=1000$, $r=0.02$, and $n=4$. Lets replace those values in our formula:
$P=1000(1+ \frac{0.02}{4} )^{4}$
$P=1020.15050$

3. Every month. In this case: $P=1000$, $r=0.02$, and $n=12$. Lets replace those values in our formula:
$P=1000(1+ \frac{0.02}{12} )^{12}$
$P=1020.18436$

4. Every week. In this case: $P=1000$, $r=0.02$, and $n=52$. Lets replace those values in our formula:
$P=1000(1+ \frac{0.02}{52} )^{52}$
$P=1020.19742$

5. Every day. In this case: $P=1000$, $r=0.02$, and $n=365$. Lets replace those values in our formula:
$P=1000(1+ \frac{0.02}{365} )^{365}$
$P=1020.20078$

We can conclude that $P$ increases as $n$ increases.