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
[tex]A=P_{0}(1 +  \frac{r}{n} )^{nt} = 1000(1+ \frac{0.02}{n})^{n} [/tex]
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