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.
n
1.) 1 (once per year)
2.) 4 (every 3 months)
3.) 12 (every month)
4.) 52 (every week)
5.) 365 (every day)

P=Po(1+ r/n) is the equation used
[o=0 placed at the bottom of the P]
Will mark as brainliest!!

Answer 1

Given:
P₀ = $1000, the principal
r = 2% = 0.02, the APR

Let the duration be 1 year 
n =  number of compounding intervals per year.
The value after 1 year is
$A=P_{0}(1 + \frac{r}{n} )^{nt} = 1000(1+ \frac{0.02}{n})^{n}$

The following table shows the results obtained from the calculator.

     n                   A
------  ------------------
     1   1020.00000
     4   1020.15050
    12   1020.18436
    52  1020.19742
  365  1020.20078