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)^n1 (once per year) a04 (every 3 months) a112 (every month) a252 (every week) a3365 (every day) a4

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

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

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

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

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

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

We can conclude that [tex]P[/tex] increases as [tex]n[/tex] increases.