Answer:

Explanation:
Interest compound is defined as

Where,
is the principal.
is the rate in decimal.
is the number of compounded periods within a year.
is the time in years.
In this case, we have

Replacing all values, we have

Now, we solve for




Our result is different because at each step we approximated results.
If you use a calculater, you would find a more exact result would be

Therefore, the right answer is the third choice.