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

So, we are going to use the compound interest formula: 
where 
is the final amount 
is the initial amount 
is the interest rate in decimal form 
is the number of times the interest is compounded per year
is the time in years 
Since in all the cases , we can omit .

1. Once per year. In this case: , , and . Lets replace those values in our formula:



2. Every three months. In this case: , , and . Lets replace those values in our formula:



3. Every month. In this case: , , and . Lets replace those values in our formula:



4. Every week. In this case: , , and . Lets replace those values in our formula:



5. Every day. In this case: , , and . Lets replace those values in our formula:



We can conclude that  increases as  increases.

+1,000)=g7e=mc2-g=7 because if you know algebra you should know g=7 to get your answer