An initial deposit of $35,000 grows at an annual rate of 7% for 23 years. Compare the final balances resulting from continuous compounding and annual compounding. We're supposed to use the equation A=P(1+r/k)^kt but I don't know what I'm supposed to use for k when the compounding is continuous. Thanks.
There is a special formula for continuous compounding
it is
Amount = Principal (e^(rt) ) where r is the annual rate compounded continuously and t is the number of years.
So for annual compounding
amount = 35000(1.07)^23 = 165 918.55
continuous compounding
amount = 35000(e^(23(.07)) = 175 098.39
thanks so much!
To compare the final balances resulting from continuous compounding and annual compounding, we can use the formula you mentioned, A = P(1 + r/k)^(kt).
In this case, r represents the annual interest rate, P is the initial deposit, t is the number of years, and k represents the compounding frequency.
For annual compounding, we use k = 1 since the compounding occurs once a year.
However, for continuous compounding, we need to use a different approach. Continuous compounding assumes that interest is calculated and added to the balance an infinite number of times within the year.
In this case, we can use the formula A = Pe^(rt), where e is the mathematical constant approximately equal to 2.71828.
Let's calculate the final balances for both scenarios:
For annual compounding:
A = P(1 + r/k)^(kt)
A = $35,000(1 + 0.07/1)^(1*23)
A = $35,000(1.07)^23
A ≈ $149,718.16
For continuous compounding:
A = Pe^(rt)
A = $35,000 * e^(0.07 * 23)
A ≈ $163,015.50
Therefore, the final balance after 23 years is approximately $149,718.16 with annual compounding and approximately $163,015.50 with continuous compounding.